Question

A pair of fair dice is cast. what is the probabiliy that the sum of the numbers falling uppermost is $9$, given that at least one of the numbers falling uppermost is a $3$?

Answer 1

Determine the probability:

A die is a six faced object that has the numbers $1\mathrm{to}6$ on its faces.

So, the numbers that can be obtained on the topmost face will be from the set $\left\{1,2,3,4,5,6\right\}$.

The number on the upper face of one of the die is $3$.

Assume that the event to obtain the sum $9$ by adding the number on the upper faces of the two dice is $A$

The number on the upper face of the second die must be $6$ for the sum to be $9$.

If the upper face of first die shows $3$, then the second die must show $6$ .

Obtaining a number on the first and the second die are independent events.

SO, the probability of each event will be multiplied.

The above mentioned result of one way of occurrence of the event $A$.

The second way of the occurrence of the event is if the upper face of second die shows $3$, then the first die must show $6$ .

The probability will be given by the sum of probabilities of each way of occurrence of the event.

Construct an expression for the probability and solve:

$\begin{array}{rcl}P\left(A\right)& =& \frac{1}{6}\times \frac{1}{6}+\frac{1}{6}\times \frac{1}{6}\\ & \Rightarrow & P\left(A\right)=\frac{1}{36}+\frac{1}{36}\\ & \Rightarrow & P\left(A\right)=\frac{1}{18}\end{array}$

Hence, the required probability is $P\left(A\right)=\frac{1}{18}$.