Coin Toss

There are $2^4=16$ possible outcomes to four coin tosses. However we are told that there is at least one head. This rules out one outcome, TTTT. So in the given problem there are fifteen possible, equally likely outcomes. Only one of these is HHHH, and so the desired probability is

$\frac{1}{15}\approx 0.067$

Let $A$ be the event that all $4$ coins show heads and let $B$ be the event that at least one of the coins shows a head. Now the conditional probability

$P(A|B) = \dfrac{P(A \cap B)}{P(B)}$ and $P(A \cap B) = P(B \cap A) = P(B|A)*P(A),$

and thus $P(A|B) = \dfrac{P(B|A)*P(A)}{P(B)}.$ (This is known as Bayes' Theorem .)

If event $A$ occurs then so must event $B,$ so $P(B|A) = 1.$

Next, since each coin can show heads with probability $\dfrac{1}{2},$ we have that $P(A) = \left(\dfrac{1}{2}\right)^{4} = \dfrac{1}{16}.$

Finally, since the complement $\bar{B}$ of $B$ is the event that all coins show tails, which occurs with probability $\dfrac{1}{16},$ we have that

$P(B) = 1 - P(\bar{B}) = 1 - \dfrac{1}{16} = \dfrac{15}{16}.$

Thus $P(A|B) = \dfrac{1*\frac{1}{16}}{\frac{15}{16}} = \dfrac{1}{15} = \boxed{0.067}$ to two significant figures.

1/15 : There are 15 of 16 outcomes that have at least one H, and only one of them is HHHH.
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