If an unbiased die,marked with -2, -1, 0, 1, | vedclass

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(B) The faces are $\{-2, -1, 0, 1, 2, 3\}$. The product is positive if no outcome is $0$ and the number of negative outcomes is even.Let $P$ be the ev

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$\frac{521}{2592}$

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(B) The faces are $\{-2, -1, 0, 1, 2, 3\}$. The product is positive if no outcome is $0$ and the number of negative outcomes is even.Let $P$ be the event of getting a positive number $\{1, 2, 3\}$,$N$ be the event of getting a negative number $\{-2, -1\}$,and $Z$ be the event of getting $0$.$P(P) = \frac{3}{6} = \frac{1}{2}$,$P(N) = \frac{2}{6} = \frac{1}{3}$,$P(Z) = \frac{1}{6}$.For the product to be positive,we must have no $0$s and an even number of negative outcomes ($0, 2, 4$ negatives).Case $1$: $0$ negatives,$5$ positives: $\binom{5}{0} (\frac{1}{2})^5 = \frac{1}{32} = \frac{81}{2592}$.Case $2$: $2$ negatives,$3$ positives: $\binom{5}{2} (\frac{1}{3})^2 (\frac{1}{2})^3 = 10 	imes \frac{1}{9} 	imes \frac{1}{8} = \frac{10}{72} = \frac{360}{2592}$.Case $3$: $4$ negatives,$1$ positive: $\binom{5}{4} (\frac{1}{3})^4 (\frac{1}{2})^1 = 5 	imes \frac{1}{81} 	imes \frac{1}{2} = \frac{5}{162} = \frac{80}{2592}$.Total probability $= \frac{81 + 360 + 80}{2592} = \frac{521}{2592}$.

If an unbiased die,marked with $-2, -1, 0, 1, 2, 3$ on its faces,is thrown five times,then the probability that the product of the outcomes is positive is:

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