Question

Two dice are thrown. What is the probability that the sum of the numbers appearing on the two dice is $11$, if $5$ appears on the first
1) $\dfrac{1}{{36}}$
2) $\dfrac{1}{6}$
3) $\dfrac{5}{6}$
4) None of these

Answer 1

Hint: First we have to write all the outcomes that we may get on rolling two dice together. Then we are to find the outcomes that have $5$ as an outcome on the first dice, and simultaneously find the probability of the event that $5$ occurs on the first dice. Then, we are to find the outcomes, in which the sum of the numbers on the two dice is $11$ with $5$ on the first dice. And, also the probability of the event that sum of numbers on dice is $11$ with $5$ on the first dice. Then, we are to use the formula of conditional probability to find the probability that the sum of the numbers appearing on the two dice is $11$, if $5$ appears on the first dice. The formula of conditional probability is, if $A$ and $B$ are two events, then,
$P\left( {\dfrac{B}{A}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}$.

Complete step-by-step solution:
Given, two dice are thrown together, then the outcomes are,

Outcomes on each dice	$1$	$2$	$3$	$4$	$5$	$6$
$1$	$\left( {1,1} \right)$	$\left( {1,2} \right)$	$\left( {1,3} \right)$	$\left( {1,4} \right)$	$\left( {1,5} \right)$	$\left( {1,6} \right)$
$2$	$\left( {2,1} \right)$	$\left( {2,2} \right)$	$\left( {2,3} \right)$	$\left( {2,4} \right)$	$\left( {2,5} \right)$	$\left( {2,6} \right)$
$3$	$\left( {3,1} \right)$	$\left( {3,2} \right)$	$\left( {3,3} \right)$	$\left( {3,4} \right)$	$\left( {3,5} \right)$	$\left( {3,6} \right)$
$4$	$\left( {4,1} \right)$	$\left( {4,2} \right)$	$\left( {4,3} \right)$	$\left( {4,4} \right)$	$\left( {4,5} \right)$	$\left( {4,6} \right)$
$5$	$\left( {5,1} \right)$	$\left( {5,2} \right)$	$\left( {5,3} \right)$	$\left( {5,4} \right)$	$\left( {5,5} \right)$	$\left( {5,6} \right)$
$6$	$\left( {6,1} \right)$	$\left( {6,2} \right)$	$\left( {6,3} \right)$	$\left( {6,4} \right)$	$\left( {6,5} \right)$	$\left( {6,6} \right)$

The total number of outcomes of the event $ = 36$
Let $A$ be the event that the first dice contains $5$.
Therefore, the suitable outcomes are, $A = \left( {5,1} \right),\left( {5,2} \right),\left( {5,3} \right),\left( {5,4} \right),\left( {5,5} \right),\left( {5,6} \right)$
Probability of $A$, that the first dice contains $5$, $P\left( A \right) = \dfrac{6}{{36}}$
Let $B$ be the event that the sum of the numbers appearing on the dice is $11$.
Therefore, the suitable outcomes are, $B = \left( {5,6} \right),\left( {6,5} \right)$
Now, the outcome common to both $A$ and $B$, $A \cap B = \left( {5,6} \right)$.
The probability of the event that the sum of the numbers is $11$, with $5$ on the first dice, $P\left( {A \cap B} \right) = \dfrac{1}{{36}}$
Now, the formula of conditional probability, that is,
$P\left( {\dfrac{B}{A}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}$.
Therefore, the probability that the sum of the numbers appearing on the two dice is $11$, if $5$ appears on the first dice $P\left( {\dfrac{B}{A}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}$
Substituting the values, we get,
$ \Rightarrow P\left( {\dfrac{B}{A}} \right) = \dfrac{{\dfrac{1}{{36}}}}{{\dfrac{6}{{36}}}}$
$ \Rightarrow P\left( {\dfrac{B}{A}} \right) = \dfrac{1}{6}$
Therefore, the probability that the sum of the numbers appearing on the two dice is $11$, if $5$ appears on the first dice is $\dfrac{1}{6}$, the correct option is 2.

Note: Many can do the mistake here by finding the probability as $P\left( {A \cap B} \right)$, but it will be just an outcome of the event, but here we are given that we are to find the probability of sum of numbers on dice is $11$, given the condition that, on the first dice the number that appeared is $5$. Hence, a condition was already given, so, we have used conditional probability. We should be careful while handling calculative steps.