Question

A die is thrown twice and the sum of the numbers appearing is observed to be 9. What is the conditional probability that the number 4 has appeared at least once?

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{3}{4}$
• D. $Noneofthese$

Answer 1

The correct option is A $\frac{1}{2}$

Let A be the event of getting the sum 9 and B be the event of getting at least one 4.
Then A = {(3, 6), (4, 5), (5, 4), (6, 3)}
B = {(1, 4), (2, 4), (3, 4) (4, 4), (5, 4), (6, 4),}
(4, 1), (4, 2), (4, 3), (4, 5), (4, 6)}
then $A\cap B=(4,5)(5,4)$
$\therefore$ Required probability $=P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P\left(A\right)}$
$=\frac{n(A\cap B)}{n\left(A\right)}=\frac{2}{4}=\frac{1}{2}$