# A dice is thrown thrown two times and the sum of the scores apperaing on the dice is observed to be multiple of 4. Then the conditional probability that the score 4 has appeared alteast once is: Option: 1 Option: 2 Option: 3 Option: 4

A = event of sum being multiple of 4

B = event that 4 has appeared atleast once

$P(A)=\frac{9}{36}=\frac{1}{4}$

$P (B)= \frac{11}{36}$

$P (A \cap B)= \frac{1}{36}$

$P \frac{B}{A}= \frac{P(A \cap B)}{P(A)}=\frac{\frac{1}{36}}{\frac{9}{36}}=\frac{1}{9}$

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