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

A dice is thrown twice

$S=(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\left(\right)2,1,(2,2),(2,3),(2,4),(2,5)(2,6)(3,1),(3,2)(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$

Here we need to find the probability that has appeared at least once. Since the sum of number oserved to be $6$

LEt$F=$ Sum of numbers is $6$

$E=4$ has appeared at least once

We need to find $P\left(E|F\right)$

$E=(1,4),(2,4).(3,4),(4,4),(5,4),(6,4),(4,1),(4,2),(4,3),(4,5),(4,6)$

$P\left(E\right)=\frac{11}{16}F=(1,5),(5,2),(2,4),(4,2),(3,3)P\left(F\right)=\frac{5}{36}$

Also$E\cap F=(2,4),(4,2)P(E\cap F)=\frac{2}{36}$

Now$P\left(E/F\right)=\frac{P(E\cap F)}{P\left(F\right)}$

$=\frac{\frac{2}{36}}{\frac{5}{36}}$

$\therefore$ required probability $=\frac{2}{5}$