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(D) Let $n$ be the number of times the dice is rolled. The probability of getting an odd number in a single throw is $p = \frac{1}{2}$,and the probabi

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$\frac{1}{2}$

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(D) Let $n$ be the number of times the dice is rolled. The probability of getting an odd number in a single throw is $p = \frac{1}{2}$,and the probability of getting an even number is $q = 1 - p = \frac{1}{2}$.The probability of getting an odd number $2$ times is given by the binomial distribution: $P(X=2) = {}^{n}C_{2} (\frac{1}{2})^{2} (\frac{1}{2})^{n-2} = {}^{n}C_{2} (\frac{1}{2})^{n}$.The probability of getting an even number $3$ times is equivalent to getting an odd number $(n-3)$ times: $P(Y=3) = {}^{n}C_{3} (\frac{1}{2})^{3} (\frac{1}{2})^{n-3} = {}^{n}C_{3} (\frac{1}{2})^{n}$.Given that $P(X=2) = P(Y=3)$,we have ${}^{n}C_{2} = {}^{n}C_{3}$.Using the property ${}^{n}C_{r} = {}^{n}C_{n-r}$,we get $2 + 3 = n$,so $n = 5$.We need to find the probability of getting an odd number an odd number of times,which is $P(X=1) + P(X=3) + P(X=5)$.$P(X=1) + P(X=3) + P(X=5) = {}^{5}C_{1} (\frac{1}{2})^{5} + {}^{5}C_{3} (\frac{1}{2})^{5} + {}^{5}C_{5} (\frac{1}{2})^{5} = \frac{1}{2^{5}} (5 + 10 + 1) = \frac{16}{32} = \frac{1}{2}$.

An ordinary dice is rolled for a certain number of times. If the probability of getting an odd number $2$ times is equal to the probability of getting an even number $3$ times,then the probability of getting an odd number for an odd number of times is

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