For each coin, there are only two possible outcomes. Either it lands on heads, or it lands on tails. The probability of a coin landing on heads is independent of other coins. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Theoretically, a fair coin

Equally as likely to land on heads or tails, so $p = \frac{1}{2} = 0.5$

10 coins:

This means that $n = 10$

What is the theoretical probability of this happening?

This is P(X = 2).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{10,2}.(0.5)^{2}.(0.5)^{8} = 0.0439$

4.39% theoretical probability of this happening

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3 years ago