A die is thrown two times and the sum of the scores appearing

Question:

A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4 . Then the conditional probability that the score 4 has appeared atleast once is :

  1. $\frac{1}{8}$

  2. $\frac{1}{9}$

  3. $\frac{1}{3}$

  4. $\frac{1}{4}$


Correct Option: , 2

Solution:

A: Sum obtained is a multiple of $4 .$

$A=\{(1,3),(2,2),(3,1),(2,6),(3,5),(4,4),$,

$(5,3),(6,2),(6,6)\}$

B : Score of 4 has appeared at least once.

$B=\{(1,4),(2,4),(3,4),(4,4),(5,4),(6,4)$ $(4,1),(4,2),(4,3),(4,5),(4,6)\}$

Required probability $=\mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)=\frac{\mathrm{P}(\mathrm{B} \cap \mathrm{A})}{\mathrm{P}(\mathrm{A})}$

$=\frac{1 / 36}{9 / 36}=\frac{1}{9}$

Leave a comment