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Let’s say that a die is rolled seven times.

Use the binomial distribution form to determine the possibility of rolling exactly five 3’s.

Please help me! I'm so confused. BTW i'm really bad at math so please explain in a really simple way. Thanks!

Maria
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    Why don't you just plug it into the formula? $P(n; k; p)= {n \choose k}p^k(1-p)^{n-k} $ where $n$ is the number of the number tries. $k$ is the exact number of successes. And $p$ is the probability of a single success.

    SO $n = 7$ rolls. $k = 5$ threes. $p = \frac 16$.

    So $P = {7 \choose 5}\frac 16^5(1-\frac 16)^{7-5}=$

     – fleablood May 09 '18 at 23:13
  • To roll a $3$ is $\frac 16$. To not roll a $3$ is $\frac 56$. To roll $5$ threes in a row is $(\frac 16)^5$. To roll not threes twice in a row is $(\frac 56)^2$. To roll five threes in a row followed by two no threes is $(\frac 16)^5(\frac 56)^2$ If you roll a three exactly 5 times and a non three exactly 2 times in 7 rolls there are ${7\choose 5}$ ways to choose which of the 7 rolls were threes and which were not. So prob to roll five threes and 2 none threes in any order is ${7\choose 5}(\frac 16)^5(\frac 56)^2$. – fleablood May 09 '18 at 23:49
  • Thank you Fleablood! I just solved it as 175/93,312 and it was correct. – Maria May 10 '18 at 00:12

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