A) \[\frac{1}{2}\]
B) \[\frac{1}{3}\]
C) \[\frac{2}{3}\]
D) \[\frac{2}{5}\]
Correct Answer: B
Solution :
Let A and B be two events such that \[A=\]getting number 2 at least once \[B=\]getting 7 as the sum of the numbers on two dice We have, \[A=\]{(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (1, 2), (3, 2), (4, 2), (5, 2), (6, 2)} and \[B=\]{(2, 5), (5, 2), (6, 1), (1, 6), (3, 4), (4, 3)} \[\therefore \] \[P(A)=\frac{11}{36},\]\[P(B)=\frac{6}{36}\] \[P(A\cap B)=\frac{2}{36}\] Now, required probability \[P(A\text{/}B)=\frac{P(A\cap B)}{P(B)}=\frac{2\text{/}36}{6\text{/}36}=\frac{2}{6}=\frac{1}{3}\]
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