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Four fair coins are tossed,what is the probability of at least getting two heads? How can I find the probability without drawing out all the results ?

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3 Answers 3

Use a binomial distribution. The probability of $k$ successes in $n$ trials, where the probability of success is $p$ is:

$$ \binom{n}{k} p^k (1-p)^{n-k} $$

Here, $p=1/2$, $n=4$, and we sum $k=2$, $k=3$, and $k=4$:

$$P = \left [ \binom{4}{2} + \binom{4}{3} + \binom{4}{4} \right ] (1/2)^4 = \frac{6+4+1}{16} = \frac{11}{16}$$

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Here's one way:

Let $A$ be the event that there are at least two heads and $B$ be the event that there are at least two tails.

Note that:

$\ \ \ $1) By symmetry, $P(A)=P(B)$.


$\ \ \ $2) $P(A\cup B)=1$.

So $$1=P(A\cup B)=P(A)+P(B)-P(A\cap B)=2P(A)-P(A\cap B);$$ thus $$P(A)=\textstyle{1\over2}\bigl(1+P(A\cap B)\bigr).$$

So now, you need only find $P(A\cap B)$, which is the probability that there are exactly two heads. This probability is $6/16$.

But, in my opinion, it's just as easy to "draw out all the possibilities".

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I think it is clever and original, but it is also not all that instructive as it only covers situations in which the symmetry exists. –  Ron Gordon Jan 7 '13 at 16:20

A -> Probability of getting atleast two heads.

P(A) = 1-Probability of getting only one head - probability of getting no heads

Probability of getting only one head = P(getting head only on the first coin + getting head only on the second coin+getting head only on the third coin+getting head only on the fourth coin) =

4*(1/2)$^4$ = 1/4

Prob of getting no heads = (1/2)$^4 $ = 1/16


P(A) = 1 - (1/4) -(1/16) = 11/16

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