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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 \frac{1}{4}
Option: 2 \frac{1}{3}
Option: 3 \frac{1}{8}
Option: 4 \frac{1}{9}

Answers (1)

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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}

Posted by

Suraj Bhandari

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