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We roll a fair $6$-sided die $5$ times. What is the probability that we get an odd number in exactly $4$ of the $5$ rolls?

To roll exactly $4$ odds out of $5$ rolls is $(1/2)^4 \cdot (1/2)$ where the odds are $1$, $3$, and $5$ for $3/6=1/2$ and the other $1/2$ is the probability of rolling an even since we want exactly $4$ odds. Since there are $5$ dice being rolled there are $\binom{5}{1}$ or $5$ ways to order it. So the probability is $5/32$. How does this look?

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    $\begingroup$ If it makes it easier to picture, note that since the probability that a single toss is odd is $\frac 12$ you might as well work with a fair coin and ask for the probability of getting $H$ in exactly $4$ out of $5$ tosses. Using the binomial distribution, that would be $\binom 54\times \left(\frac 12\right)^5=\frac 5{32}$, just as you computed. $\endgroup$ – lulu Oct 16 '17 at 21:42
  • $\begingroup$ So HHHHT, HHHTH,HHTHH, HTHHH, THHHH. Or 5 out of 32 possiblities. $\endgroup$ – ddswsd Oct 16 '17 at 21:44
  • $\begingroup$ That is exactly right. $\endgroup$ – lulu Oct 16 '17 at 21:48
  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Oct 16 '17 at 23:46

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