Question

A die is thrown twice and the sum of the numbers appearing is observed to be $6.$ What is the conditional probability that the number $4$ has appeared at least once ?

Answer 1

Given: A die is thrown twice
Then, sample space is
$S=\left\{\begin{array}{cccccc}(1,1),& (1,2),& (1,3),& (1,4),& (1,5),& (1,6),\\ (2,1),& (2,2),& (2,3),& (2,4),& (2,5),& (2,6),\\ (3,1),& (3,2),& (3,3),& (3,4),& (3,5)& (3,6),\\ (4,1),& (4,2),& (4,3),& (4,4),& (4,5),& (4,6),\\ (5,1),& (5,2),& (5,3),& (5,4),& (5,5),& (5,6),\\ (6,1),& (6,2),& (6,3),& (6,4),& (6,5),& (6,6)\end{array}\right\}$

Total no of elements in $S=6\times 6=36$
Let A: sum of numbers is $6\&$
$B:4$ has appeared at least once
$A=\left\{\right(1,5),(5,1),(2,4),(4,2),(3,3\left)\right\}$
$P\left(A\right)=\frac{5}{36}$ ......$\left(1\right)$
Now,
$B:4$ has appeared at least once
$B=\left\{\begin{array}{c}(1,4),(2,4),(3,4),(4,2),(4,3),(5,4),(4,5),\\ (4,4),(6,4),(4,6),(4,1)\end{array}\right\}$
$P\left(B\right)=\frac{11}{36}$ .....$\left(2\right)$
Also,
$A\cap B=\left\{\right(2,4),(4,2\left)\right\}$
$P(A\cap B)=\frac{2}{36}$ .....$\left(3\right)$

We know that,
$P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P\left(A\right)}$ ....$\left(4\right)$

Substituting values from $\left(1\right)\&\left(3\right)$ in $\left(4\right).$
$P\left(\frac{B}{A}\right)=\frac{\frac{2}{36}}{\frac{5}{36}}$

$\therefore P\left(\frac{B}{A}\right)=\frac{2}{5}$
The conditional probability that the number $\left(4\right)$ has appeared at least once , given that sum of numbers is six $=\frac{2}{5}$