What’s the probability of getting 3 heads and 7 tails if one flips a fair coin 10 times

Thursday, March 7, 2019

Your question is related to the binomial distribution.

You do

n = 10

. The probability of one successful trial is

p = \frac{1}{2}

You want

k = 3

successes and

n - k = 7

failures. The probability is:

(\binom{n}{k}) p^{k} (1 - p)^{n - k} = (\binom{10}{3}) \cdot {(\frac{1}{2})}^{3} \cdot {(\frac{1}{2})}^{7} = \frac{15}{128}

One way to understand this formula: You want

k

successes (probability:

p^{k}

) and

n - k

failures (probability:

(1 - p)^{n - k}

). The successes can occur anywhere in the trials, and there are

(\binom{n}{k})

to arrange

k

successes in

n

trials.

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