Bill rolls a fair number cube 20 times. What is the probability that 10 of his results are multiples of 3?

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Bill rolls a fair number cube 20 times. What is the probability that 10 of his results are multiples of 3?

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Answer 1 · 2018-04-29T12:06:09

$\approx 0.0543$ $~~0.0543$

Explanation:

We can do this using a binomial probability.

$n \sum k = 0 C_{n, k} {(p)}^{k} {(1 - p)}^{n - k} = 1$ $sum_(k=0)^(n)C_(n,k)(p)^k(1-p)^(n-k)=1$

For this question, $n = 20, k = 10$ $n=20, k=10$

With a roll of a fair number cube, see that the multiples of 3 are 3 and 6, giving a probability of rolling a multiple of 3 on any given roll as $p = \frac{1}{3}$ $p=1/3$ .

This gives:

$C_{20, 10} {(\frac{1}{3})}^{10} {(\frac{2}{3})}^{10}$ $C_(20,10)(1/3)^10(2/3)^(10)$

$184756 \times {(\frac{1}{3})}^{10} \times {(\frac{2}{3})}^{10} \approx 0.0543$ $184756xx(1/3)^10xx(2/3)^10~~0.0543$

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Bill rolls a fair number cube 20 times. What is the probability that 10 of his results are multiples of 3?

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