Question

Two dice are thrown. Given the information that the number on the first die iseven, and the number on the second is $\mathbf{<}\mathbf{4}$, set up an appropriate sample space andanswer the following questions.

(a) What are the possible sums and their probabilities?

(b) What is the most probable sum?

(c) What is the probability that the sum is even?

Short answer

(a) The possible sums and their probabilities are.$\begin{array}{cccccccc}Sum& 3& 4& 5& 6& 7& 8& 9\\ probability& \frac{1}{9}& \frac{1}{9}& \frac{2}{9}& \frac{1}{9}& \frac{2}{9}& \frac{1}{9}& \frac{1}{9}\end{array}$

(b) The most probable sumis 5 and 7.

(c) The probability that the sum is evenis $\frac{1}{3}$.

Step-by-step solution

Definition of Non-Uniform Sample Space

Non-Uniform Sample space of any experiment is the set of all possible mutually exclusive events such that each point has different probability.

(a) Creation of thesample space for experiment when two dice are rolled

When two dice are rolled, there are 9 points in the sample space.

Find the sample space for the given condition.

$\left(\begin{array}{ccc}2,1& 2,2& 2,3\\ 4,1& 4,2& 4,3\\ 6,1& 6,2& 6,3\end{array}\right)$

Find the possible sum of the sample space with associated probabilities when two die are rolled.

$\begin{array}{cccccccc}Sum& 3& 4& 5& 6& 7& 8& 9\\ probability& \frac{1}{9}& \frac{1}{9}& \frac{2}{9}& \frac{1}{9}& \frac{2}{9}& \frac{1}{9}& \frac{1}{9}\end{array}$

(b) Determination ofthe most probable sum

It can be observed from the obtained sample space and their probabilities that 5 and 7 have the probability of $\frac{2}{9}$and so, they are the most probable sum.

Thus, the most probable sum is 5 and 7.

(c) Determination of the probability that the sum is even

The obtained sample space for sum contains each sum possible.

Find the probability of getting an even sum by adding all the associated probabilities where sum is even that is for 4, 6, and 8.

$\begin{array}{c}\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9}\\ =\frac{1}{3}\end{array}$

Thus, the probability that the sum is even is .$\frac{1}{3}$