Probability - Solved Examples - Number Cube

This set of solved examples on probability deals with the simple probability problems based on a number cube.

Solved Example 1

What is the probability that on tossing a number cube, you get the number 4?

Solution

Step 1: Find the total number of possible outcomes

On tossing a number cube, you can get either one of the six faces up. Thus, there are six possible outcomes.

Step 2: Find the number of favorable outcomes

The number 4 is present only on one face on a number cube. Thus the number of favorable outcomes is 1.

Step 3: Apply the formula for theoretical probability

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

`P(4) = 1/6`

Thus, the probability of getting a 4 on rolling a number cube is `1/6`.

Solved Example 2

What is the probability of getting an even number on rolling a number cube?

Solution

Step 1: Find the total number of possible outcomes

On tossing a number cube, you can get either one of the six faces up. Thus, there are six possible outcomes.

Step 2: Find the number of favorable outcomes

Out of the six numbers on a number cube (1 through 6), three are even (2, 4 and 6). Thus the number of favorable outcomes is 3.

Step 3: Apply the formula for theoretical probability

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

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

Thus, the probability of getting an even number on rolling a number cube is `1/2`.

Solved Example 3

What is the probability of getting a number greater than or equal to 5 on rolling a number cube?

Solution

Step 1: Find the total number of possible outcomes

On tossing a number cube, you can get either one of the six faces up. Thus, there are six possible outcomes.

Step 2: Find the number of favorable outcomes

There are only two numbers, 5 and 6, greater than or equal to 5 on a number cube. Thus the number of favorable outcomes is 2.

Step 3: Apply the formula for theoretical probability

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

`P("greater than or equal to 5") = 2/6 = 1/3`

Thus, the probability of getting a number greater than or equal to 5 on rolling a number cube is `1/3`.

Solved Example 4

What is the probability of getting a number lesser than 4 on rolling a number cube.

Solution

Step 1: Find the total number of possible outcomes

On tossing a number cube, you can get either one of the six faces up. Thus, there are six possible outcomes.

Step 2: Find the number of favorable outcomes

There are three numbers lesser than 4 on a number cube (1, 2 and 3). Thus the number of favorable outcomes is 3.

Step 3: Apply the formula for theoretical probability

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

`P(4) = 3/6 = 1/2`

Thus, the probability of getting a 4 on rolling a number cube is `1/2`.

Solved Example 5

What is the probability of getting an odd number on rolling a number cube?

Solution

Step 1: Find the total number of possible outcomes

On tossing a number cube, you can get either one of the six faces up. Thus, there are six possible outcomes.

Step 2: Find the number of favorable outcomes

Out of 1 to 6, three numbers (1, 3 and 5) are odd on a number cube. Thus the number of favorable outcomes is three.

Step 3: Apply the formula for theoretical probability

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

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

Thus, the probability of getting an odd number on rolling a number cube is `1/2`.

Solved Example 6

What is the probability of getting a prime number on rolling a number cube?

Solution

Step 1: Find the total number of possible outcomes

On tossing a number cube, you can get either one of the six faces up. Thus, there are six possible outcomes.

Step 2: Find the number of favorable outcomes

Out of 1 through 6, there are three prime numbers: 2, 3 and 5. Note that 1 is not a prime number. Thus, there are three favorable outcomes.

Step 3: Apply the formula for theoretical probability

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

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

Thus, the probability of getting a prime number on rolling a number cube is `1/2`.

Solved Example 7

What is the probability of getting a number greater than 7 on rolling a number cube?

Solution

Step 1: Find the total number of possible outcomes

On tossing a number cube, you can get either one of the six faces up. Thus, there are six possible outcomes.

Step 2: Find the number of favorable outcomes

A number cube contains numbers 1 to 6. It does not have the number 7 or any number greater than 7. Thus, the number of favorable outcomes is zero.

Step 3: Apply the formula for theoretical probability

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

`P("greater than 7") = 0/6 = 0`

Thus, the probability of getting a number greater than 7 on rolling a number cube is 0.

Solved Example 8

What is the probability of getting a composite number on rolling a number cube?

Solution

Step 1: Find the total number of possible outcomes

On tossing a number cube, you can get either one of the six faces up. Thus, there are six possible outcomes.

Step 2: Find the number of favorable outcomes

A number cube contains numbers 1 to 6, out of which the composite numbers are 4 and 6. Thus, there are two composite numbers, and hence two favorable outcomes.

Step 3: Apply the formula for theoretical probability

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

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

Thus, the probability of getting a composite number on rolling a number cube is `1/3`.

Can all quadratic functions be written in intercept form?

The simple answer is: No, not all quadratic functions can be written in the intercept form. Only those quadratic functions can be written in the intercept form that have real solutions.

Let us understand this further.

What is intercept form?

The intercept form of a quadratic equation is as follows:

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

where 'p' and 'q' are the roots of the quadratic function. We know that a quadratic function has 'roots' or 'zeros' only when its graph touches or intersects the x-axis.

Functions that can be written in intercept form

Some quadratic functions have real solutions; That is, their graphs touch or intersect the x-axis at some point/s. There graphs look like either of the two graphs given below:

Graph touches the x-axis. This type of quadratic function has only one root and its intercept form is of the form `y = a(x - p)^2`

Graph intersects the x-axis at two points, say at x = p and x = q; This type of quadratic function has intercept form `y = a(x - p)(x - q)`

Such quadratic functions, that have a real solution, can be written in intercept form. This is because the roots or zeros (or real solutions) of the quadratic function are the values of 'p' and 'q' in the intercept form.

Functions that can't be written in intercept form

On the other hand,some quadratic functions do not have any real solution. Their graph does not touch or intersect the x-axis at any point. Their graphs look like either one of these graphs:

Graph does not touch or intersect the x-axis and faces up. This function can't be written in intercept form since it does not have real roots.

Graph does not touch or intersect the x-axis and faces down. This function can't be written in intercept form since it does not have real roots.

These quadratic functions can not be written in the intercept form because we don't have any real number to put for in place of 'p' and 'q' in their intercept form.

Note on complex/imaginary roots

When a quadratic function does not have any real root, it can have imaginary roots. For example, the function `y = x^2 + 1` does not have any real roots, but it's imaginary roots are `i` and `-i`. Thus, if we put `p = i` and `q = -i` in the intercept form, we can write the function as follows:

`y = (x - i)(x + i)`

Thus, if we involve complex numbers (or imaginary numbers), all quadratic functions can be written in intercept form. On the other hand, if we consider only real numbers, a quadratic function can only be written in intercept form if it has one or two real roots.

Factor Theorem

The Factor Theorem says that

If a function/expression becomes zero at a value 'c' then `x - c` is a factor of that function/expression.

For example, `x^2 - 3x + 2` becomes zero at x = 2:

`(2)^2 - 3*(2) + 2 = 4 - 6 + 2 = 0`

Thus, by factor theorem, `x - 2` is a factor of the expression `x^2 - 3x + 2`. Another value at which this expression becomes zero is x = 1. Thus `x - 1` is also a factor of the expression. Thus, we can write

`x^2 - 3x + 2 = (x - 1)(x - 2)`

Prime Expressions

An algebraic expression is called 'prime' when it can not be factored. For example, 5x + 1 is a prime expression because there is no common factor between the two terms '5x' and '1' which can be factored out.

We know that a prime number has only two factors - 1 and itself. Similarly a prime expression has only two factors - 1 and itself. No other number, term or expression is its factor.

Conversely, `5x + 10` is not a prime expression because the two terms in it, '5x' and '10' have a common factor 5, which can be factored out of the expression, thus making it `5(x + 2)`.

Linear expressions (the ones in the examples above) can be factored only by one method: factoring out a common factor, but quadratic and higher degree expressions can be factored by various other methods. Thus, in order to test whether a higher degree expression is prime or not, you need to try to factor it in different ways.

A sure-fire signal that an expression (quadratic and higher degree) can not be factored is when its graph does not have any x-intercepts. This is a direct result of the factor theorem.

More on Prime and Non Prime Quadratic Expressions

Odds against/in favor of an event

"Odds against" an event is the ratio of the probability of not happening of that event to the probability of happening of that event.

For example, odds against getting a particular number, say 5, on rolling a number cube is the ratio of the probability of not getting 5 to the probability of getting 5 on rolling that number cube.

"Odds in favor of" an event is the ratio of the probability of happening of that event to the probability of not happening of that event.

For example, odds in favor of getting the number 5 on rolling a number cube is the ratio of the probability of getting the number 5 to the probability of not getting the number 5.

Thus, we conclude that the odds against an event are the opposite of the odds in favor of the event.

Let us understand the meaning of "odds" with an example. This will also lead us on to the formula written below it.

What are the odds in favor of getting the number 5 on rolling a number cube?

First, calculate the probability of getting the number 5 on rolling a number cube. We use the formula for theoretical probability to calculate it.

`P(5) = "Number of favorable outcomes"/"Total number of outcomes"`

`P(5) = 1/6`

Now calculate the probability of not getting the number 5 on rolling the number cube. This is the complement of the probability of getting 5, thus,

`P("not 5") = 1 - P(5) = 1 - 1/6 = 5/6`

Now, odds in favor of getting the number 5 on rolling a number cube are given by the ratio of the above probabilities.

`"Odds in favor of getting 5" = (P(5))/(P("not 5"))`

`"Odds in favor of getting 5" = (1/6)/(5/6) = 1/5"`

Thus, the odds in favor of getting the number 5 on rolling a number cube are `1/5`.

From the above example, we can conclude that the formula for odds in favor of an event can be written as follows:

`"Odds in favor of an event E" = (P("E"))/(P("not E"))`

Similarly, since odds against an event is the ratio of the probability of not happening of that event to the probability of happening of that event, therefore its formula is

`"Odds against of an event E" = (P("not E"))/(P("E"))`

Thus, as stated previously, we can say that the odds against an event are the complete opposite of the odds in favor of it. In other words, the odds against an event are the reciprocal of the odds in favor of an event. For example, if the odds against an event are `1/2`, then the odds in favor of that event are `2/1`.

The above example leads us to another conclusion: Since the odds against or in favor of an event are the ratio of probabilities and not the probabilities themselves, therefore they can be greater than 1as opposed to probability of an event (recall that the probability of an event can not be greater than 1).

Further, if we use the formula for theoretical probability, we can write

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

and,

`P("not E") = "Number of outcomes that are not favorable"/"Total number of outcomes"`

If we place the above two formulas for P(E) and P(Not E) in the formula for odds in favor of an event E, we get

`"Odds in favor of event E" = (P("E"))/(P("Not E"))`

`"Odds in favor of event E" = ("Number of favorable outcomes"/"Total number of outcomes")/("Number of outcomes that are not favorable"/"Total number of outcomes")`

Simplifying, we get,

`"Odds in favor event E" = "Number of favorable outcomes"/"Number of outcomes that are not favorable"`

The above formula is extremely useful in calculating the odds in favor of an event. It helps you calculate the odds in favor of an event without calculating the probability of its happening or not happening.

Since the odds against an event are the reciprocal of the odds in favor of it, thus, we can also write the formula for odds against an event as follows:

`"Odds against an event E" = "Number of outcomes that are not favorable"/"Number of favorable outcomes"`

Solved Examples

1. There are five red, four blue and three white marbles in a bag. What are the odds against and in favor of getting a red marble on drawing one marble from the bag?

There are five red marbles in the bag, therefore the number of favorable outcomes for drawing a red marble is 5. The other 7 marbles are not red. Thus the number of outcomes that are not favorable are 7. Thus,

`"Odds against getting a red marble" = "Number of outcomes that are not favorable"/"Total number of outcomes"`

`"Odds against getting a red marble" = 7/5`

Since the odds in favor of an event are the opposite of the odds against it, therefore,

`"Odds in favor of getting a red marble" = 1/"Odds against it" = 5/7`

2.  A number cube is rolled. What are the odds against getting a number greater than 4 on it? What are the odds in favor?

There are two numbers, 5 and 6, greater than 4 on a number cube. Thus, the number of favorable outcomes is 2. The other four numbers on the number cube (1 to 4) are not greater than 4. Thus the number of outcomes that are not favorable is 4.

`"Odds against getting a number greater than 4" = "Number of outcomes that are not favorable"/"Number of outcomes that are favorable"`

`"Odds against getting a number greater than 4" = 4/2 = 2/1`

Since the odds in favor of an event are the opposite of the odds against an event, therefore

`"Odds in favor of getting a number greater than 4 = 1/2"`

3. One card is drawn from a deck of fifty two playing cards. What are the odds in favor of getting a red King?

There are two red Kings in a standard deck of fifty two cards. Thus the number of favorable of favorable outcomes is 2. The other 50 cards are not red Kings. Thus the number of outcomes that are not favorable are 50.

`"Odds in favor of getting a red King" = "Number of favorable outcomes"/"Total number of outcomes"`

`"Odds in favor of getting a red King" = 2/50= 1/25`

4. It rained on four out of five consecutive days in a week. What are the odds against raining on the sixth day?

Since it rained on four days, number of favorable outcomes for raining = 4
Since it did not rain on one day, therefore number of non-favorable outcomes = 1
Applying the formula for "odds against",

`"Odds against raining" = "Number of non-favorable outcomes"/"Number of favorable outcomes" = 1/4`

5. Team A won three of the four matches against Team B. What are the odds in favor of team A winning the fifth match?

Since team A won three matches, therefore number of favorable events for wining a match of team A = 3
Since team A did not win one match out of four, therefore number of non-favorable outcomes for winning a match of team A = 1
Applying the formula for "odds in favor",

`"Odds in favor of team A winning the match" = "Number of favorable outcomes"/"Number of non-favorable outcomes" = 3/1 = 3`

Hyperbolic Functions

Hyperbolic functions are defined as follows

• `cosh(theta) = 1/2(e^x + e^(-x))`
• `sinh(theta) = 1/2(e^x - e^(-x))`
• `tanh(theta) = (sinh(theta))/(cosh(theta)) = (e^x - e^(-x))/(e^x + e^(-x))`
• `coth(theta) = (cosh(theta))/(sinh(theta)) = (e^x + e^(-x))/(e^x + e^(-x))`
• `sech(theta) = 1/(cosh(theta)) = 2/(e^x + e^(-x))`
• `csch(theta) = 1/(sinh(theta)) = 2/(e^x - e^(-x))`

Identities of hyperbolic functions

• `cosh^2(x) - sinh^2(x) = 1`
• `cosh^2(x) + sinh^2(x) = cosh(2x)`
• `cosh(2x) = cosh^2(x) - 1`
• `cosh(2x) = sinh^2(x) - 1`
• `sinh(2x) = 2sinh(x)cosh(x)`
• `sinh^(-1)(x) = log(x + sqrt(x^2 + 1))`
• `cosh^(-1)(x) = log(x + sqrt(x^2 - 1))` 

Derivatives of hyperbolic functions

• `d/dx sinh(x) = cosh(x)`
• `d/dx cosh(x) = sinh(x)`
• `d/dx tanh(x) = sech^2(x)`
• `d/dx coth(x) = -csch^2(x)`
• `d/dx sech(x) = -sech(x)tanh(x)`
• `d/dx csch(x) = -csch(x)coth(x)`

Probability - Combinations Formula Application

Introduction

The combinations formula is very useful in calculating probability in many problems. It helps you easily calculate the number of favorable events and the total number of events in the sample space. Thus it saves you the time of manually listing the sample space and counting the number of total and number of favorable events.

Formula

The formula for combinations is as follows.

`C_r^n = {n!}/{r!(n-r)!}`

This formula helps you calculate the number of favorable outcomes and the total number of outcomes in the given problem, which then helps you calculate the probability. The solved examples below will help you understand how this formula is applied to probability problems.

Solved Examples

Problem 1

There are three red, five blue and two orange marbles in a bag. Two marbles are selected from it without replacement. What is the probability that both the marbles will be red?

In this problem we will use the combinations formula to calculate the number of favorable outcomes and the total number of outcomes. The number of favorable outcomes are all those outcomes in which two red marbles are selected. Since there are a total of three red marbles in the bag, hence the different ways in which two red marbles can be selected from three red marbles are given by

`C_2^3 = (3!)/(2!(3-2)!) = 3`

Therefore the number of favorable outcomes is 3. Now we will calculate the total number of outcomes. Total number of ways in which two balls (of any color) can be selected from 10 balls is given by

`C_2^{10} = (10!)/(2!(10-2)!) = 45`

Now that we have the number of favorable and the total number of outcomes, we calculate the probability by the help of the formula for theoretical probability.

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

`P("two red marbles") = 3/45 = 1/15`

Therefore the probability of selecting two red marbles out of the bag is `1/15`.

Problem 2

There are twenty boys and ten girls in a class. What is the probability that two of the selected students will both be boys.

Calculate the total number of possible outcomes. The total number of different ways in which two students can be selected from a total of 20 + 10 = 30 students are given by

`C_2^{30} = (30!)/(2!(30 - 2)!) = 435`

Now calculate the number of favorable outcomes. Since a favorable outcome is selecting two boys and there are twenty boys in the class, hence the total number of favorable outcomes is given by

`C_2^{20} = (20!)/(2!(20-2)!) = 190`

Therefore the total number of possible outcomes is 435 while the number of favorable outcomes is 190. Applying the formula for theoretical probability,

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

`P("two boys are selected") =  190/435 = 38/87`

Problem 3

Out of five hundred students in a school, two hundred opted to study Math, two hundred opted to study English and one hundred opted to study Biology. No student can opt for more than one subject. What is the probability that five student chosen at random will all be studying Math?

Find the total number of possible outcomes. There are a total of 500 students and the different number of ways in which 5 are chosen are

`C_5^{500} = (500!)/(5!(500-5)!) = 255244687600`

Find the number of favorable outcomes. Choosing five students all studying math will give a favorable outcome. Since 200 students opted for math, therefore the different number of ways in which five students can be selected from the 200 students is given by

`C_5^{200} = (200!)/(5!(200-5)!) = 2535650040`

Now applying the formula for theoretical probability,

`P("all five students study math") =  2535650040/255244687600 = 0.01`

Problem 4

There are ten apples in the box out of which five are green and five are red. What is the probability that two apples selected from the box will have one red and one green apple?

Find the total number of possible outcomes. The total number of ways of selecting two apples from ten are given by

`C_2^{10} = (10!)/(2!(10-2)!) = 45`

There are five red and five green apples. The number of ways of selecting one red apple out of five are given by `C_1^5=5` and those of selecting one green apple out of five are given by `C_1^5 = 5`. Therefore the total number of ways of selecting one red and one green apple (by the fundamental principle of counting) are `5 times 5 = 25`. Thus, total number of favorable outcomes is 25.

Now applying the formula for theoretical probability,

P(one red and one green) =  `"Number of favorable outcomes"/"Total number of outcomes"`

P(One red and one green) = `25/45 = 5/9`

Therefore the probability that the two apples selected will be one green and one red is `5/9`.

Problem 5 

Seventy people were surveyed for their favorite outdoor games. Twenty said football, thirty said baseball, ten said basketball and ten said lawn tennis. What is the probability that three people selected from the seventy people will all like basketball?

Find the total number of possible outcomes. The total number of different ways of selecting three out of seventy people are given by

`C_3^{70} = (70!)/(3!(70-3)!) = 54740`

Find the number of favorable outcomes. A total of ten people play basketball. The total number of different number of ways of selecting three people out of the ten are given by

`C_3^{10} = (10!)/(3!(10-7)!) =120`

Now apply the formula for theoretical probability,

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

P(all three people like basketball) = `120/54740 = 6/2737`

Thus the probability that all the three people selected will like basketball is `6/2737`.

Probability - Binomial Theorem

Introduction

• The Binomial Theorem can be used to find probability.
• It is commonly used when the number of experiments is large.
• In this method, you don't have to list the sample space of the experiment.
• This method helps you calculate the probability that an event 'a' will occur a specific number of times when the experiment is repeated a specific number of times.
• For example, if a coin is tossed 1000 times, using the Binomial Theorem, you can calculate the probability that P(heads) will occur exactly 450 times.

Formula

`P(x) = ^nC_r(a)^x(1-a)^{n-x}`

In the above formula,

• 'a' is the probability of occurrence of the favorable event (given in the question)
• '1 - a' means the probability of non-occurrence of the favorable event
• 'n' is the number of times the experiment is repeated
• P(x) means the probability of an event occurring 'x' times.

Use

The binomial theorem helps you to calculate probability that an event 'a' will occur a given number of times when an experiment (that can result in 'a') is performed a given number of times. For example, if you toss a coin a thousand times, the binomial theorem can help you get the probability of getting exactly 470 heads out of 1000 tosses.

Probability of A and B

Definition

P(A and B) simply means "The probability of occurrence of both events, A, and B."

For example, if a coin is tossed and a number cube is rolled together, the probability of getting heads on the coin and the number 4 on the cube is represented by P(heads and 4).

P(A and B) is also written as `P(A \cap B)` and P(AB).

Formula

There are two formulas to find P(A and B) depending on whether A and B are independent or dependent events.

If events A and B are independent,

`P("A and B") = P(A) * P(B)`

If the probability of event A does not change whether event B occurs or not, and vice-versa, then A and B are called independent events. In such case, the probability of occurrence of both A and B is the product of their individual probabilities.

On the other hand, if events A and B are not independent,

`P("A and B") = P(A) * P(B|A)`

Events A and B are said to be dependent if the probability of either event changes depending on whether the other event has occurred or not. In such a case, the probability of occurrence of both A and B together is given by the above formula.

`P(B|A)` is the conditional probability of B given that event A has occurred. To understand it, read this post on conditional probability

In addition to the above formulas, there is also a method to find P(A and B) by simple counting techniques. This is shown in the example below.

Solved Examples

Problem 1

Two coins are tossed together. What is the probability that you get heads on the first coin and tails on the second.

First determine whether the two events are independent or dependent. Since the outcome of of tossing one coin does not affect the outcome of tossing the other coin, hence the two events are independent.

Find the probabilities of each event.

P(heads on first coin) = `1/2`
P(tails on second coin) = `1/2`

Now use the appropriate formula to calculate the probability of occurrence of both events. Since the two events are independent, we will use the following formula.

P(A and B) = P(A) * P(B)

P(heads on first and tails on second) = P(heads on first) * P(tails on second)

P(heads on first and tails on second) = `1/2 * 1/2 = 1/4`

Problem 2

A coin and a number cube is tossed together. What is the probability that you get heads on the coin and the number 3 on the number cube?

Determine whether the given events are dependent or independent. Since the outcome of tossing the coin does not affect the outcome of rolling the number cube in any form, therefore the two events are independent.

Find the individual probabilities of both events.
P(heads on coin) = `1/2`
P(3 on cube) = `"Number of 3's on a number cube"/"Total number of faces on a number cube" = 1/6`

Apply the appropriate formula to calculate the probability of both events occurring.

P(heads on coin and 3 on cube) = P(heads on coin) * P(3 on cube) = `1/2 * 1/6 = 1/12`

Therefore the probability of getting a heads on tossing a coin and 3 on rolling a number cube is `1/12`.

Problem 3

There are two bags of marbles. One bag contains three blue and four white marbles while the other bag contains two red and two green marbles. One marble is selected from each bag. What is the probability that you get a blue marble from the first bag and a red marble from the second?

Determine whether the two events are independent or dependent. Since the outcome of selecting a marble from the first bag does not affect the outcome of selecting a marble from the second bag, hence the two events are independent.

Find the individual probabilities of both events.
P(blue from first bag) = `"Number of blue marbles in the first bag"/"Total number of marbles in the first bag" = 3/7`
P(red from second) = `"Number of red marbles in the second bag"/"Total number of marbles in the second bag" = 2/4 = 1/2`
P(blue from first and red from second) = `3/7 * 1/2 = 3/14`

Problem 4

Two number cubes are tossed. What is the probability that you get an even number on the first cube and an odd number on the second?

The two given events are independent since the outcome of rolling the first cube does not affect that of rolling the second cube. Thus,

P(even on first and odd on second) = P(even on first) * P(odd on second)

There are a total of six numbers on a number cube out of which three are even and three are odd. Thus,

P(even on first) = `"Number of even numbers on a cube"/"Total number of numbers on a number cube" = 3/6 = 1/2`

P(odd on first) = `"Number of odd numbers on a cube"/"Total number of numbers on a number cube" = 3/6 = 1/2`

Applying the formula above,

P(even on first and odd on second) = `1/2 * 1/2 = 1/4`

Problem 5 

From a standard deck of 52 cards, you pick two cards. What is the probability that the first card is a red Ace and the second is a black Queen?

Before drawing the first card, there are a total of 52 cards and a total of two red aces in it. Thus,

P(red ace) = `2/52 = 1/26`

If the first card is a red ace, then there are a total of 51 and two black queens left in the deck. Thus,

P(black queen) = `1/51`

Since the two events above are independent, therefore,

P(red ace and black queen) = P(red ace) * P(black queen) = `1/25 * 1/51 = 1/1326`

Therefore the probability of getting a red ace and then a black queen on drawing two cards from a deck of 52 cards is `1/1326`

Experimental Probability

Definition

Experimental probability is defined as the probability calculated on the basis of the results of an experiment.

For example, we know that on tossing a coin, the probability of getting heads is `1/2`. This is calculated theoretically and is thus called theoretical probability. Suppose you toss a coin a thousand times, and you get heads 470 times and tails 530 times. Then based on the results of this experiment, the probability of getting heads is `470/1000`. Since `470/1000` is the probability calculated from the results of an experiment, we call it "experimental probability". 

Further, we note that `1/2` is not equal to `470/1000`. Thus experimental probability is not always equal to the theoretical probability.

Formula

`P(A) = "Number of times you get the favorable result"/"Number of times the experiment is performed"`

Solved Examples

Problem 1

Out of ten running contests, Ginny got the first position in 8. What is the probability that she will get the first position in the next running contest?

Total number of running contests = 10

Number of contests in which Ginny got the first position = 8

Thus, applying the formula for experimental probability,

`P("first position in next contest") = "Number of contests in which Ginny got the first position"/"Total number of running contests"`

`P(first position in next contest") = 8/10 = 4/5`

Problem 2

On rolling a dice, a student got an even number six out of the ten times that he rolled the dice. What is the probability that he will get an even number on throwing the dice now?
  

Total number of times the dice is rolled = 10

Number of times that an even number occurred = 6

Thus, applying the formula for experimental probability,

`P(even number) = "Number of times that an even number occurred"/"Total number of times the dice is rolled"`

`P(even number) = 6/10 = 3/5`

Problem 3

In the last 30 days it rained on 20 days. What is the probability that it will rain today?

Number of days on which it rained = 20
Total number of days observed = 30

Thus, applying the formula for experimental probability,

`P("it will rain on next day") = "Number of days on which it rained"/"Total number of days observed"`

`P(it will rain on the next day") = 20/30 = 2/3`

Problem 4

Dan missed the school bus three out of five days a week. What is the probability that he will miss the school bus today?

Number of times Dan missed the school bus = 3
Total number of days = 5

Applying the formula for experimental probability,

`P("Dan will miss the bus today") = "Number of times he missed the school bus"/"Total number of days"`

`P("Dan will miss the bus today") = 3/5`

Problem 5 

Mrs. Sandy told Jimmy to get 12 eggs from the dairy farm. On the way home, Jimmy broke 5 on the twelve eggs. Then Mrs. Sandy told Jimmy to get one more egg. What is the probability that Jimmy will not break the egg this time?

Total number of eggs = 12
Number of eggs broken = 5

By the formula for experimental probability, the probability that Jimmy will break an egg is

`P("egg breaks") = 5/12`

The two events, that is, "the egg will break" and "the egg will not break" are complimentary events. Thus, the sum of their probabilities is 1.

The probability that Jimmy will not break an egg is

`P("egg does not break") = 1 - 5/12 = 7/12`

Exercise

Seventy eight out of a hundred students passed in an examination.