Probability: Dealing with a Deck of Cards

There are a number of questions of probability that can be formed on a pack of cards. First, a little intro about a pack of cards:

Standard Deck of 52 Cards

There are a total of 52 cards in a pack of cards, divided into two colors:

• Red (26 cards)
• Black (26 cards)

There are a total of 4 suits of cards, two each in red and black. They contain 13 cards each. They are:

• Spades (black) - 13 cards
• Clubs (black) - 13 cards
• Diamonds (red) - 13 cards
• Hears (red) - 13 cards

Each of these four suits, having 13 cards, contains the following cards:

• 1 King
• 1 Queen
• 1 Jack
• 1 Ace
• 2 to 10 numbered cards

Face cards: Each suit has a King, a Queen and a Jack. These three cards are known as the face cards of each suit. Since there are four suits, and three face cards in each suit, so there are a total of 4 * 3 = 12 face cards in a deck of 52 cards.

It would be logical to relate to the standard deck (or pack) of fifty two cards explained above whenever you encounter the phrase "the deck", or "the pack" in the discussion below.

Probability of Getting a Particular Card

What do you think might be the probability of getting a particular card, say a King of Hearts, when you draw one card out of a standard deck of fifty two cards?

It is simple. There are a total of fifty two cards to draw from, so the total number of different possible outcomes is fifty two. There is just one King of Hearts in a standard deck of fifty two cards. Hence the total number of favorable outcomes is one. So, applying the formula for theoretical probability:

`P("outcome") = "Number of favorable outcomes"/"Total number of possible outcomes"`

`P("King of Hearts") = 1/52`

(If you didn't understand this one, you probably need to understand the basics of calculating theoretical probability. These are discussed in these posts:

• Basic Terms in Probability
• Solving Basic Probability Problems

So better read that before proceeding on.)

Another one: What is the probability of getting an Ace when drawing one card out of a standard deck of fifty two cards?

This one is a little different from the previous one. There are four aces in a standard deck of fifty two cards, one of each of the suits. We already know, from the previous question's solution, that there are a total of fifty two possible outcomes when you draw a single card out of a deck of fifty two cards. The number of favorable outcomes is, however, different in this question. There are four aces in a pack of cards, so if you draw any one of the four aces, you get a favorable outcome. Hence the number of favorable outcomes is four. Applying the formula for theoretical probability,

`P("outcome")="Number of favorable outcomes"/"Total number of outcomes"`

`P("Ace")=4/52=1/13`

Hence the probability of drawing an ace from a pack of 52 cards is 1/13, which means, that out of every thirteen cards you draw one by one, one of them is probably an ace.

What is the probability of getting a red Ace or a black Jack on drawing a card from a standard deck of fifty two cards?

Again the total number of possible events is fifty two, since there are fifty two cards in a standard deck of cards. However, the number of favorable events is different in this one. How many red aces and black jacks can you find in a pack of fifty two cards? Please refer to the section "Standard Deck of 52 Cards" above to try to get that yourself before reading on. If you need help, however, we state again that there are a total of four aces in a standard deck, each one of the four different suits. The four suits are spades and clubs (black) and hearts and diamonds (red). So there are two red aces and two black aces in a deck.

Similarly there are four Jacks in a deck of cards, each one of the four different suits. Likewise, there are two red Jacks and two black Jacks in a deck.

The question asks you to find the probability of getting a red ace or a black jack on drawing one card. Since there are two red aces and two black Jacks in a deck of cards and you can draw any one of them to get the favorable outcome, hence the number of favorable events is four (= 2 red aces + 2 black jacks).

Applying the formula for theoretical probability again, we get

`P("outcome")="Number of favorable outcomes"/"Total number of outcomes"`

`P("red Ace or black Jack")=4/52 = 1/13`

Till now you must have understood how to calculate the probability of getting a particular card or one of a particular combination of cards from a standard deck of fifty two cards. So let us advance to a few problems on probability that deal with drawing two cards from the deck.

Click here to see more solved problems on probability of selecting a particular type of card from a deck.

Drawing Two Cards Without Replacing

This is a typical question with special regards to "without replacing" the first card. This means that you draw the second card without replacing the first card in the pack. So there are fifty one cards left to draw from after drawing the first card, and to draw the second card you have to choose from these fifty one cards. After drawing the second card, fifty cards are left in the deck.

What is the probability of drawing two red Aces one by one when drawing two cards out of a deck of fifty two cards?

Consider a deck of fifty two cards lying in front of you. There are a total of fifty two cards in the deck containing two red Aces. The first card that you draw has to be a red ace. On drawing the first card, there are a total number of fifty different possible outcomes possible and a total of two favorable outcomes from which one can occur. So, by the formula for theoretical probability, the probability of getting a red ace is:

`P("first card - red ace") = 2/52 = 1/26`

Suppose you did draw a red ace (against all odds). Then there is only one other red ace left in the deck of cards in front of you. The deck also now contains only fifty one cards, since you have drawn one card and did not replace it. Probability of getting a red ace in drawing the second card is:

`P("second card - red ace") = "Number of red aces"/"Total number of cards"`

`P("second card - red ace") = 1/51`

Consider: The probability of drawing a red ace in the first draw is 1/26 and that of drawing another red ace in the second draw is 1/51. What is the probability that both of these events can occur together? Here comes the concept of independent events, and their compound probability. The compound probability, that is, the probability of both of the above events happening together, is a product of their individual probabilities:

`P("both cards - red ace")=1/26 * 1/51 = 1/1326`

Hence the probability of getting a red ace in both the first and second draws is 1 in 1326 double draws. Quite a large number (or, equivalently, quite a small probability), isn't it?

What is the probability of getting two cards of the same color when drawing them one by one without replacing from a standard deck of fifty two cards?

There are cards of two colors in a deck: red and black. There are 26 red and 26 black cards in it. So you can either draw two red cards or two black cards from the pack.

First consider the probability of getting two red cards: Out of fifty two cards, 26 are red, so the probability of getting a red card when drawing your first card is `26/52 = 1/2`. After drawing the first card, there are 51 cards left in the pack, out of which, 25 are red and 26 are black because the first card you drew was red in color. Thus, to draw a red card from the remaining set of cards, the probability of drawing a red card is `25/51`.
The two events of drawing the cards one by one are two independent events and thus the probability of their occurring together is given by the product of their individual probabilities. Thus the probability of getting two red cards is `1/2 * 25/51 = 25/102`

Now let us consider the probability of getting two black cards: Out of fifty two cards, 26 are black, so the probability of getting a black card when drawing your first card is `26/52 = 1/2`. After drawing the first card, there are 51 cards left in the pack, out of which, 25 are black and 26 are red because the first card you drew was black in color. Thus, to draw a black card from the remaining set of cards, the probability of drawing a black card is `25/51`.
The two events of drawing the two cards one by one are two independent events and thus the probability of their occurring together is given by the product of their probabilities. Thus the probability of getting two black cards is `1/2 * 25/51 = 25/102`

P(both cards of same color) = P(both red or both black) = P(both red) + P(both black)
P(both cards of same color) = `\frac{25}{102} + \frac{25}{102} = \frac{50}{102} = \frac{25}{51}`

Thus the probability of selecting two cards of the same color from a deck of cards is `25/51`.

Greatest Common Divisor (GCD)

Introduction

Consider any two numbers, such as 12 and 36. Make a list of all the numbers which are factors (numbers which can divide 12 and 36 without leaving any remainder) of both 12 and 36: {1, 2, 3, 4, 6, 12}. The greatest of these factors is '12'. Thus, 12 is called the greatest common divisor (GCD) or the greatest common factor (GCF) of 12 and 36.

Thus, we can define GCD as the largest number which is a factor (or divisor) of both the given numbers.

Let's consider another example. List all numbers which are factors of both 18 and 27: {1, 3, 9}. Clearly, 9 is the largest number which can divide both 18 and 27 without leaving any remainder (that is, it is a factor of 18 and 27). Thus, 9 is the GCD of 18 and 27.

Prime factorization method to find GCD

It's quite easy to find the GCD of small numbers such as given above. On the other hand, when you have large numbers, finding their GCD is simpler by using the prime factorization method. Let us learn the method by taking an example: 512 and 216.

Prime factorize both numbers

`512 =   2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2`

`216 =   2 \times 2 \times 2 \times 3 \times 3 \times 3`

To find their GCD, take all the prime factors of both the numbers which are common to both numbers and multiply them. In the above prime factorization, both the numbers 512 and 216 have three 2's common, but no other number/s are matching. Thus, their GCD is,

GCD = `2 \times 2 \times 2 = 8`

Thus, the greatest common divisor of 512 and 216 is 8.

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Expansion Formulas for Factoring

Probability: Independent and Dependent Events

In probability, two events are said to be independent if they do not affect the probability of each other. For example, you toss two coins one after the other. The probability of getting heads on the first coin is 1/2 and that of getting tails on the second coin is 1/2. Even if you toss only one coin, or even three coins, the probability of getting either heads or tails on each coin does not change. It remains 1/2. These kind of events which do not affect the probability of each other are called independent events.

In order to understand it better, let us now discuss two events that are not independent. Consider a bag full of marbles of different colors. The bag has 5 blue marbles, 10 white marbles and 4 red marbles. Suppose you take two marbles from the bag in two consecutive draws. Suppose you drew a blue marble on the first draw. Then, after this first draw, you are left with 18 marbles (4 blue, 10 white and 4 red). Now the probability of getting a white marble in the second draw is equal to

`P("white marble on second draw") = "Number of white marbles"/"Total number of marbles"`

`P("white marble on second draw") = 10/18 = 5/9`

(If you did not understand this, it would be better if you read this post first)
Thus the probability of getting a white marble on the second draw is 5/9. But this is when the first draw gave you a blue marble. What if you got, instead of a blue marble, you got a white marble in the first draw? Will the probability of getting the white marble on the second draw remain the same? Let's see.

Suppose you draw a white marble on the first draw. Then, after the first draw, there are a total of 18 marbles in the bag out of which 5 are blue, 9 are white and 4 are red. Now, the probability of getting a white marble in the second draw is

`P("white marble on second draw") = "Number of white marbles"/"Total number of marbles"`

`P("white marble on second draw") = 9/18 = 1/2`

Notice that we got two different results for the probability of getting a white marble on the second draw here: 5/9 and 1/2. When the first marble drawn is blue, the probability of getting a white marble on the second draw is 5/9, and when the first marble drawn is white, the probability of getting a white marble on the second draw is 1/2. Thus, we see that the probability of getting a white marble in the second draw is dependent on what you get in the first draw. Thus, the probability of the second event is dependent on the outcome of the first event. These type of events are called dependent events and are not independent events.

Sample Space, Favorable Outcomes and Formula for Theoretical Probability

Theoretical probability, as discussed earlier, is calculated without even performing the experiment. It is calculated based on the sample space and the number of favorable outcomes out of that sample space. What is sample space then?

Sample Space

Sample space is the collection of all possible outcomes or results from an experiment. It is thus useful to list the sample space of an experiment, because it helps you calculate the total number of possible outcomes of an experiment. The following examples of sample space will illustrate it clearly.

Favorable Outcomes

In simple words, a "Favorable Outcome" is the desired result of an experiment. For example, you toss and coin and want heads, then getting heads is a favorable outcome, whereas getting tails is not a favorable outcome.

Example:

Find the favorable outcomes to getting an even number when rolling a dice.

A dice has six faces, numbered 1 through 6. The even numbers on it are 2, 4 and 6. Thus, there are three favorable outcomes out of total six possible outcomes when rolling a dice.

Formula for Theoretical Probability

Theoretical probability is calculated solely on the basis of the sample space and the number of outcomes in the sample space that meet the condition required in any given question. Its formula is:

`P("outcome") = "Number of Favorable Outcomes"/"Total Number of Possible Outcomes"`

The "outcome" written in parenthesis after 'P' represents the particular outcome for which the probability is being calculated. For example, if you are calculating the probability of getting the number 3 when throwing a dice, then you would write 3 in place of "outcome", that is, you will write P(3).

The number of favorable outcomes in the above formula are the same as discussed above, that is, the number of outcomes out of the sample space that match the given condition in the question.

The total number of possible outcomes is the total number of outcomes listed in the sample space. For example, for throwing two dice, the total number of possible outcomes is 36.

It would serve better to further your knowledge of probability by solving, or seeing how they are solved, some simple questions on probability (that apply the above formula).

Solving Simple Probability Problems

Prev: Probability

We'll start with solving a few simple problems based on theoretical probability. Solving these problems includes counting the sample space, number of favorable outcomes and using the formula for theoretical probability. So you will learn these three fundamental terms related to probability along the way.

The first problem, that of getting heads when a coin is tossed, has already been discussed in the previous post. So here we will start with another one:

What is the probability of getting two heads when you throw a coin twice in a row?

The probability of getting two heads twice in a row is 1/4. How did we get that? This we will discuss below:

While solving any problem related to probability, you first need to compute two things:

• Total number of events in the sample space
• Number of favorable outcomes

The sample space is a collection of all the possible outcomes that may arise from the given event. Here the event is throwing a coin twice in a row, and the outcomes can be:

Throw 1	Throw 2
Heads	Tails
Heads	Heads
Tails	Heads
Tails	Tails

As you can see, there are a total of four possible outcomes listed above. These are nothing but a combination of all the possible outcomes in each throw. Hence the total  number of outcomes in the sample space is equal to four.

Now we have to calculate the total number of favorable outcomes. Favorable outcomes are those outcomes in the sample space the probability of which we are to calculate. In this question we have to calculate the probability of getting two heads. From the table above, we can see that out of four possible combinations, only one has two heads. Hence the number of favorable events is one.

Now applying the formula for theoretical probability,

`P("outcome") = "Number of favorable outcomes"/"Total number of outcomes"`

Plug the values into the formula to get

`P("two heads") = 1/4`

Hence, the probability of getting two heads is 1/4 when a coin is thrown twice in a row. Note that if, instead of throwing one coin twice in a row, you had thrown two coins, it would have made no difference; the probability would remain same since the sample space and number of favorable events will not change. You can try to solve it that way as well.

From the above discussion you must have got some idea of what a sample space is, what are favorable outcomes, and, how is the formula for theoretical probability used. These basic terms will become clearer to you as you go through the next set of problems:

Another problem:

What is the probability of getting the same face on both the throws of a coin?

This question asks you to find the probability of getting two heads or two tails when you throw a coin twice in a row.

Refer to the table above, it is clear that the sample space is same for this question. Hence the total number of outcomes in the sample space is 4.

Again from the table above, you can see that out of the four possible outcomes, we have one outcome in which both the throws result in heads and one in which both the throws result in tails. Thus, in total, there are 2 outcomes in which both the throws result in the same face. Hence the total number of favorable events is 3.

Applying the formula for theoretical probability,

`P("outcome") = "Number of favorable outcomes"/"Total number of outcomes"`

Substituting the values of number of favorable outcomes and total number of outcomes,

`P("same face on both") = 2/4 = 1/2`

Hence the probability of getting the same face is 1/2, when a coin is thrown twice. Again, if instead of throwing one coin two times, you had thrown two coins once, the sample space and probability would remain the same as above.

Too many throws of a coin! Let us let the coin have rest for some time. Now advancing to other types problems:

What is the probability of getting the number 2 when a dice is thrown?

As you may know, a dice has six faces. The faces are numbered 1 to 6. So whenever you throw a dice, you can get one of the six faces showing up. Hence the total number of outcomes possible is equal to 6.

Only one face on a dice shows the number 2. Hence the number of favorable outcomes is 2.

Applying the formula for theoretical probability, we get:

`P("outcome") = "Number of favorable outcomes"/"Total number of outcomes"`

`P(2) = 2/6 = 1/3`

Next question:

What is the probability of getting an even number on throwing a dice?

As before, the total number of possible outcomes in the sample space is equal to six.

However the number of favorable outcomes is equal to three, since out of the six numbers (1 to 6) on the six faces of a dice, three are even (2, 4 and 6) and three are odd (1, 3 and 5). Since the question is for even numbers, hence there are three possible favorable events.

Applying the formula for theoretical probability,

`P("outcome") = "Number of favorable outcomes"/"Total number of outcomes"`

`P("even number") = 3/6 = 1/2`

Hence the probability of getting an even number of throwing a dice is 1/2.

You may now be quite clear about calculating theoretical probability of simple experiments as discussed above in the question. You can also refer to this page to understand the definitions of the common terms used in probability, that is, the sample space, the favorable events and the formula for theoretical probability.

Probability

Probability in math is the numerical representation of the chance of happening or not happening of an event. If I ask you, "What is the probability of getting a heads on flipping a coin?", you may answer "It is 1/2". Do you know how you got 1/2? Maybe not. If not, then let me explain:

There are two sides of a coin: A heads and a tails. When you flip a coin, you can get either heads or tails. So there are two possible 'outcomes' to the 'event' of throwing a coin (notice the terms outcomes and event). Out of these two possible outcomes, getting heads is just one. So the probability of getting heads becomes 1/2.

From the above discussion, we can conclude that probability of a particular outcome from an event is the number of outcomes that you want over the total number of outcomes that you can have. In a formula, this would appear as:

`P("outcome") = "Number of outcomes you want"/"Total number of outcomes"`

This is the formula for Theoretical Probability. Theoretical probability is called 'theoretical' because when we calculate it, we find out the probability before even performing the task; Do you have to throw the coin to know that the probability of getting heads is 1/2? No. You know that before you throw the coin. In other words, you calculated the probability without performing the task or experiment. This is Theoretical Probability.

Now let us actually perform an experiment. Let us take a fair coin (a fair coin means a coin that is not biased towards showing either heads or tails) having two sides, heads and tails, and throw it a thousand times (yes, you read that correct, a thousand times) and record the number of heads and tails we get.

Instead of going into the details of performing the experiment, let us see the results here:

• Number of tails = 495
• Number of heads = 505

We know that the probability of getting a heads is 1/2, but the above results show that out of one thousand throws, the coin showed heads 505 times, which is not exactly half of the number of trials. How can that be possible?

This is possible because Theoretical Probability is, well, theoretical. It is not always in accordance with the actual results of an experiment. If we were to determine the probability from the experiment above, then we would say that the probability of getting heads is 505 out of 1000, that is 505/1000 = 0.505 (which, again, is very near to 0.5). This probability is based on the actual results of the experiment and holds only for the experiment that you performed. It may not hold good for any other similar experiment. This kind of a probability is called Experimental Probability.

In dealing with probability problems, generally we do not perform the experiments stated in the problems to compute the probabilities, and hence, we generally only compute the Theoretical Probability in solving them.

Probability - Throwing Two Dice

As explained earlier, there are a total of 36 possible outcomes when you throw two dice once, and the outcome of each unique pair of numbers, for instance, 1 on the first and 2 on the second dice, is 1/36.

Different types of questions on probability can be based on throwing two dice. Mainly, these are:

• Probability of getting a particular sum
• Probability of getting a sum greater/lesser than a particular value
• Probability of getting particular numbers on both dice
• Problems involving conditional probability

Each of these question types is discussed below:

Getting a particular sum

Suppose, on throwing the two dice, the first dice shows the number 4 while the second one shows the number 6. The sum of these two numbers is 10. How many other such possible pairs of numbers are possible on the two dice to get a sum of 10? These are given in the table below:

dice 1	dice 2	Sum
4	6	10
6	4	10

dice 1	dice 2	Sum
2	6	8
3	5	8
4	4	8
5	3	8
6	2	8

Sum	Number of favorable outcomes
2	1
3	2
4	3
5	4
6	5
7	6
8	5
9	4
10	3
11	2
12	1

Thus, only two outcomes out of the possible thirty six outcomes show a sum of 10. This means that total number of outcomes favorable is two.  Hence we can say that the probability of getting a sum of 10 on the throw of two dice is:

`P("sum of 10") =  "number of favorable outcomes"/"total number of possible outcomes"`

`P("sum of 10")  = 2/36 = 1/18`

Let us consider another sum, eight. How many pairs of numbers can you come up with, both between 1 and 6, that add up to eight? The following table shows the numbers:

There are a total of five outcomes which give you a number of eight. Hence total number of favorable outcomes in this case is five. Probability of getting a sum of eight is given by:

`P("sum of 8") =  "number of favorable outcomes"/"total number of possible outcomes"`

`P("sum of 8")  = 5/36`

Similarly you can come up with the probabilities of getting other sums ranging from 2 to 12 (since 2 is the least possible sum and 12 is the greatest). A patter is obvious from the table below which specifies the total number of pairs of numbers on two dice that gives each sum. The number of outcomes increases from 1 to 6 and then decreases to 1 again from the sum of 7. 

Getting sum greater than or lesser than a given number

What is the probability of getting a sum greater than or equal to ten on the throw of two dice?

When throwing two dice, total number of pairs of numbers that can come up is 36. Out of these 36 outcomes, total number of outcomes which give a sum of 10 is 3 (refer to the table above), total number of outcomes which give a sum of 11 is 2 and total number of outcomes which give a sum of 12 is 1. Hence total number of outcomes which give a sum of 10 or greater is 3 + 2 + 1 = 6. Thus the number of outcomes favorable is 6. By using the theoretical probability formula, we get

`P("outcome") = "number of favorable outcomes"/"total number of possible outcomes"`

`P("sum greater than or equal to 10")= 6/36 = 1/6`

Getting particular numbers on both the dice

What is the probability of getting a 2 on the first dice and a 5 on the second one?

As is clear from the section on throwing a single dice above, the probability of getting any given number on a single throw of a dice is equal to 1/6. Hence the probability of getting a 2 on the first dice is 1/6 and that of getting a 5 on the second dice is 1/6 as well. But what is the probability of getting this combination of numbers, 2 and 5, on the two dice respectively, out of all the other possible pairs of numbers possible?

Notice that whatever number you get on the first dice, the probability of getting a particular number on the second dice does not change. It always remains 1/6. Thus the two outcomes - one, throwing the first dice, and two, throwing the second dice - are independent of each other.

Whenever you have to find the probability of two independent outcomes occurring together, you multiply their respective probabilities.

Thus the probability of getting 2 on the first dice and 5 on the second dice is equal to

`P("2 on 1st and 5 on 2nd dice") = 1/6 * 1/6 = 1/36`

There is another way of solving the above problem. Out of the 36 possible outcomes when throwing two dice, only one outcome is such that a 2 shows up on the first dice and a 5 shows up on the second dice. Hence by the formula of theoretical probability:

`P("outcome") =  "number of favorable outcomes"/"total number of possible outcomes"`

`P("2 on 1st and 5 on 2nd dice") = 1/36`

Modifying this question a bit, what is the probability of getting the pair of numbers 2 and 5 when throwing two dice?

Now this question asks us to find the probability of getting a 2 and a 5 when throwing two dice; it does not tell us whether we expect to get a 2 on the first dice and 5 on the second dice or 2 on the second dice and 5 on the first dice. Thus, we have to take into account both the cases.

We know that there are 36 possible outcomes when two dice are thrown. Out of these 36 possible outcomes there is only one outcome in which a 2 shows up on the first dice and a 5 shows up on the second dice. Furthermore, there is another outcome in which you get 5 on the first dice and 2 on the second dice. Thus, there are two favorable outcomes which give us the pair of numbers 2 and 5.

By the formula for theoretical probability:

`P("outcome") =  "number of favorable outcomes"/"total number of possible outcomes"`

`P("2 and 5") = 2/36 = 1/18`

This question can be extended to getting particular number sets on both dices, which is given in the worksheet at the end.

Problems Involving Conditional Probability

What is the probability of getting an even number on the second dice if it is given that the first dice shows up the number 4?

In conditional probability the total number of possible outcomes is limited by some constraint. Here the constraint is that the first dice shows up a 4 necessarily. Out of the 36 possible outcomes possible in the throw of two dice, the first dice shows up 4 in only six outcomes. Hence the total number of possible outcomes here is lessened from 36 to just 6. These are:

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

Now out of these six outcomes, how many does the second dice show an even number on it?

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

Easily, the highlighted ones above show an even number on the second dice. There are three such possible outcomes in which the second dice shows a six on it while the first one shows the number 4. Hence the total number of favorable outcomes are 3.

Thus, by the formula of theoretical probability, we get:

`P("outcome") = "Number of favorable outcomes"/"Total number of outcomes"` 

`P("even number on second dice"/"first dice shows 4") = 3/6 = 1/2`

The major thing to keep in mind for these type of questions is that normally you would have solved it like this (this is the incorrect way but is given below to clarify the use of conditional probability here)

Probability of getting a four on the first dice and an even number on the second dice is equal to

`P("outcome") = "Number of favorable outcomes"/"Total number of outcomes"`

Number of favorable outcomes is three since only three outcomes are there in which the first number is 4 and the second number is even: (4, 2), (4, 4) and (4, 6) 

Total number of outcomes is 36 since you get 36 different number pairs on throwing two dice. 

So

`P("4 on first dice and even on second dice") = 3/36 = 1/12`

The above method of solving it seems correct but it is incorrect since we did not use conditional probability here. The problem specifically states that it is "given that the first dice shows a 4". Whenever there is a statement like this one, in which the problem specifically tells you of one condition, the question is generally based on conditional probability.

Now, what is the meaning of conditional probability? Conditional probability is the probability of one outcome given that the another outcome yielded some specific outcome given in the problem. In the problem discussed above, the specific outcome given is that you get a four on the first dice. This implies that you have to narrow the sample space to include only those outcomes in which the given condition is satisfied. That is what we did in the first approach: We narrowed the total number of possible outcomes, which is 36 for two dice, to only six outcomes in which the number four shows up on the first dice. Hence, in the approach involving conditional probability, the total number of possible outcomes is narrowed down to include the given specific condition.

These where the main types of problems of probability that can be made on throwing two dice. If you have any other question on it, please post it in the comments. If it seems alright then we may try to include it in this or another post.

Thank you.

Quadratic Equation with 'p' and 'q'

A quadratic equation in 'p' and 'q' is a factored quadratic equation which is written as follows:

`y = a(x - p)(x - q)`

In the above quadratic equation, the variables are y and x whereas the constants are a, p and q, that is, in place of a, p and q you will generally find numbers there.

For example, the equation

`y = (x + 2)(x - 3)`

is a quadratic equation in which a = 1, p = -2 and q = 3

Characteristics of a quadratic equation with p and q

A quadratic equation with p and q has the following characteristics:

• 'p' and 'q' are the x-intercepts, the roots or what you could also call the solutions of the quadratic equation
• 'a' is a constant value which tells you a number of things about the quadratic equation. These are discussed below

p and q are the x-intercepts or solutions

p and q are the x-intercepts or solutions of the quadratic equation. For example, consider the quadratic equation (x + 4)(x - 4) = 0. Its solutions are -4 and 4 because the numbers in place of p and q in this equation are -4 and 4.

Now consider the following quadratic function

`y = (x + 2)(x - 3)`

Its graph is shown below. It clearly has x-intercepts at x = -2 and x = 3.

p and q are the x-intercepts

It must however be kept in mind that the sign of p and q are opposite to the sign which appears in the equation. For example, when you compare the equation

`y = (x + 2)(x - 3)`

with the equation

`y = a(x - p)(x - q)` 

you get

`x + 2 = x - p`
`=> 2 = -p`
`=> p = -2`

and you also get

`x - 3 = x - q`
`=> -3 = -q`
`=> q = 3`

Clearly, the values of p and q have opposite signs to the numbers in their places in the equation. Remembering this helps you make less errors while solving such quadratic equation problems.

Importance of 'a' 

The 'a' in the quadratic equation `y = a(x - p)(x - q)` plays an important role as it defines more than one characteristics of the parabola, which are covered one by one below:

Facing up/down or right/left

A parabola faces upwards or right when the value of 'a' is positive. The following images show the graphs of two quadratic equations in which the value of `a` is positive (a is equal to 1 in both the graphs)

Graph of y = (x + 1)(x - 1) where a = 1
Parabola facing up

Graph of x = (y + 1)(y - 1) where a = 1
Parabola is facing right

On the other hand, when 'a' has a negative value, the parabola, that is, the graph of the quadratic equation, faces down or left as shown in the graphs below:

Graph of y = -(x + 1)(x - 1) where a = -1
Parabola is facing left

Graph of x = -(y + 1)(y - 1) where a = -1
Parabola is facing down

Calculating the length of latus rectum

The value of 'a' is useful in obtaining the length of the latus rectum and other things such as coordinates of focus, equation of directrix and equation of axis of the parabola (For more information on the latus rectum, focus, directrix and axis, please visit this post). Let us first consider the latus rectum. For any value of 'a' in a quadratic equation, the length of the latus rectum is equal to four times the value of 'a'. This is explained in detail here.

Calculating the distance of the focus from the vertex, or the distance of the vertex from the directrix

In general, the distance of the focus from the vertex of a parabola is equal to 'a'. Furthermore the perpendicular distance of the directrix from the vertex is also equal to 'a'.

Probability - Throwing One Dice

A number of questions can be framed on this:

• Probability of getting an even/odd number
• Probability of getting a number greater/lesser than 3
• Probability of getting an even number greater than 3

Solved Examples

What is the probability of getting an even number on throwing a dice once?

There are six numbers on a dice: 1, 2, 3, 4, 5, and 6. Hence total number of outcomes possible on throw of a dice once are six (any one of the six numbers can show up). There are three even numbers on a dice: 2, 4 and 6. Hence total number of outcomes favorable to us are three (since any one of these three even numbers can show up). Hence by applying the formula for theoretical probability, we get,

`P("outcome") = "number of favorable outcomes"/"total number of possible outcomes"`

`P("even number") = 3/6 = 1/2`

Hence the probability of getting an even number when throwing a dice once is 1/2. Similarly, the probability of getting an odd number on a single throw of a dice is 1/2.

What is the probability of getting a number greater than  4 on a single throw of a dice?

There are six numbers on a dice: 1, 2, 3, 4, 5, and 6. Hence total number of outcomes possible on throw of a dice once are six (any one of the six numbers can show up). There are two numbers greater than four on a dice: 5 and 6. Hence total number of outcomes favorable to the outcome are two. Hence by applying the formula for theoretical probability, we get,

`P("outcome") = "number of favorable outcomes"/"total number of possible outcomes"`

`P("number greater than 4") = 2/6 = 1/3`

Hence the probability of getting a number greater than 4 when throwing a dice once is 1/3. Similarly other questions can be framed on it. You can solve the exercise for more practice. The above two questions can be combined to form another question:

What is the probability of getting an even number greater than 3 on a single throw of a dice?

Again we will use the same procedure in solving the problem. There are six numbers on a dice: 1, 2, 3, 4, 5, and 6. Hence total number of outcomes possible on throw of a dice once are six (any one of the six numbers can show up). There are three numbers greater than three on a dice: 4, 5 and 6. Out of these three number, there are two even numbers: 4 and 6. Hence total number of outcomes favorable to the outcome are two.
Hence by applying the formula for theoretical probability, we get,

`P("outcome") = "number of favorable outcomes"/"total number of possible outcomes"`

`P("even number greater than 3") = 2/6 = 1/3`

Hence the probability of getting an even number greater than 3 when throwing a dice once is 1/3.

Worksheet

Q.No.	Question	Answer
1	What is the probability of getting the number 6 on a single throw of a dice?	1/6
2	What is the probability of getting the number 7 on a single throw of a dice?	1/7
3	What is the probability of getting an odd number on a single throw of a dice?	1/2
4	What is the probability of getting a number greater than 2 on a single throw of a dice?	2/3
5	What is the probability of getting a number lesser than or equal to three on a single throw of a dice?	2/3
6	What is the probability of getting an odd number lesser than 5 on a single throw of a dice?	1/3
7	What is the probability of getting an even number greater than or equal to 5 on a single throw of a dice?	0

Probability - Throwing Dice

Throwing a single dice

A number of interesting, and simple, questions can be framed on throwing a single dice. As you may be aware, a dice has six faces numbered one to six.

A dice

A dice has six faces numbered one to six

When you throw a single dice once, either one of the six faces can come up. Thus when you throw a dice the total number of possible outcomes is six. The probability of, say, number 1 coming up is given by the formula for theoretical probability:

`P(1) = "number of favorable outcomes"/"total number of possible outcomes"`

Number of favorable outcomes is equal to one because there is only one face of the dice that can show number 1 on it. Total number of possible outcomes is six because there are six faces on a dice and any one of the six faces can show up. Hence the probability of the number one showing up on throwing a dice is given by:

`P(1) = 1/6`

Thus, when you throw a dice, the probability of any one of the six numbers showing up is equal to 1/6.

Different types of questions based on this are explained in: "Probability - Throwing One Dice"

Throwing Two Dice

Throwing two dice or throwing a single dice two times, both are the same things. The probabilities of different outcomes are same in both cases. Here we shall consider throwing two dice for simplicity.

On throwing two dice, you can get many pairs of two numbers on the two dice. For example, 1 on the first dice and 3 on the second. How many different pairs as such are possible? A total of 36 combinations are possible on throw of two dice. This can be understood as follows:

Suppose you get the number 1 when you throw the first dice. Now you throw the second dice. You can get any one of the six numbers on the second dice. Thus, for each number on the first dice you can get six numbers on the second dice. Since there are six numbers on the first dice, hence total number of possible pairs of numbers you can get are 6 + 6 + 6 + 6 + 6 + 6 = 36.

These possible pairs are shown below:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

Thus we have one important information already for solving the questions based on throw of two dice: total number of possible outcomes is equal to 36.

Different types of questions based on probability when throwing two dice are explained: "Throwing Two Dice"

Throwing Three Dice

If you throw three dice computation of the probabilities become a bit more tedious since the sample space, that is, the total number of possible combinations of numbers on the three dice, increases tremendously (it becomes 6 * 6 * 6 = 216). Hence, rather than using the theoretical probability approach that involves counting the number of favorable outcomes, here we use another approach that involves the binomial theorem. So if you are not familiar with the binomial theorem please read this post: Application of Binomial Theorem in Probability