In a single throw of two dice , find the probability that neither a doublet nor a total of 8 will appear
Solution
Correct Answer: Option B
n(S) = 36
A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
B = { (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) }
\(n\left(A\right)=6,\;n\left(B\right)=5,\;n\left(A\cap B\right)=1\)
∴ Required probability = \(P\left(A\cup B\right)\)
= \(P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)\)
= \(\frac6{36}+\frac5{36}-\frac1{36}\)
= \(\frac5{18}\)
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