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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.

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