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A die is rolled five times. What is the probability that exactly two of the results are equal to three?

We are taught to draw the trees and calculate it this way, however for a fair die being rolled five times, the diagram would be very messy. Is there another way to solve this?

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    $\begingroup$ This is a straight-out-of-the-box application of the binomial distribution. $\endgroup$ Nov 5, 2017 at 1:23
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    $\begingroup$ Wiki link to the binomial distribution. Included in the link are examples. $\endgroup$
    – JMoravitz
    Nov 5, 2017 at 1:27
  • $\begingroup$ Can you refer me to how it cant be applied? or where I can see it? thanks! $\endgroup$
    – Lola1984
    Nov 5, 2017 at 1:27
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    $\begingroup$ The wiki link includes the example about an unfair coin and getting a particular number of heads out of a certain number of flips. Instead of asking about getting a particular number of heads, here we are asking about getting a particular number of "threes." The example is exactly like this one with the only differences being flavor and the number of times the die is rolled (or coin is flipped). $\endgroup$
    – JMoravitz
    Nov 5, 2017 at 1:47
  • $\begingroup$ The diagram’s not that messy. You only have two possibilities that you care about at each branch point: whether or not a three was rolled, so the tree will have $2^5=32$ leaves, which isn’t all that big to me. $\endgroup$
    – amd
    Nov 5, 2017 at 17:41

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So you're looking for 2 successes (rolling a 3) from 5 trials with a probability of success of 1/6.

\begin{equation} P(success) = {5 \choose 2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{5-2} \end{equation}

\begin{equation} P(success) = 10\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^3 \end{equation}

P(2 threes from 5 rolls) = 0.1608 to 4 decimal places.

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