Question

A pair of dice is rolled, what is the probability that they sum to 7 given that neither of dices shows a 2?

Answer 1

Hint: We first find the combinations where the dices give a sum of 7 given that neither of the dices shows a 2. We find the number of outcomes and also find the total event. Then we use the theorem of probability where $ p\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)} $ . We find the solution putting the values in the equation.

Complete step by step solution:
A pair of dice is rolled. We have to find the probability that they sum to 7 given that neither of the dice shows a 2.
We first find the outcomes which in summation form gives 7.
For every dice the range of outcomes are $ 1,2,3,4,5,6 $ .
So, the formations for sum to be 7 is $ 1+6,2+5,3+4 $ . There are 3 such formations which yields a sum of 7.
Now it’s given that neither of the dice shows a 2 which means we have to omit the combination of $ 2+5 $ .
The remaining are $ 1+6,3+4 $ .
We now denote the event of rolling the dice as event S where $ n\left( S \right)=6\times 6=36 $ . Also, we denote the event of rolling the dices with sum being 7 given that neither of dices shows a 2 as event A where $ n\left( A \right)=2\times 2=4 $ .
The multiplication of 2 and 2 happened as the permutation of the numbers $ 1+6,3+4 $ for two dices as $ 1+6,3+4,6+1,4+3 $ .
Now we have to find the probability of event A which is $ p\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)} $ .
So, \[p\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{4}{36}=\dfrac{1}{9}\].
The probability that they sum to 7 given that neither of dices shows a 2 is \[\dfrac{1}{9}\].
So, the correct answer is “\[\dfrac{1}{9}\]”.

Note: We need to be careful about the combination of the numbers. The choices $ 1+6,3+4 $ were the combinations of numbers. We multiplied the number 2 for the armament of the dice for the numbers.