\(\frac{1}{8}\)
\(\frac{1}{24}\)
\(\frac{1}{12}\)
\(\frac{1}{4}\)
Correct answer is C
Chance of x = \(\frac{1}{2}\)
Change of Y = \(\frac{2}{3}\)
Chance of Z = \(\frac{1}{4}\)
Chance of Y and Z only occurring
= Pr (Y ∩ Z ∩ Xc)
where Xc = 1 - Pr(X)
1 - \(\frac{1}{2}\) = \(\frac{1}{2}\)
= Pr(Y) x Pr(Z) x Pr(Xc)
= \(\frac{2}{3}\) x \(\frac{1}{4}\) x \(\frac{1}{2}\)
= \(\frac{1}{12}\)
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