If 5 coins are flipped what is the probability of getting only one head?

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How would I do this question? I know if the question said: at least one head then I would do:

${5\choose0}=1$

$2^5=32-1 = 31 $

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3 Answers

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We assume that the coin is fair and is flipped fairly.

There are $2^5$ equally likely strings of length $5$ made up of the letters H and/or T.

There are precisely $5$ strings that have exactly $1$ H and $4$ T.

So the required probability is $\dfrac{5}{2^5}$.

Remark: Suppose that a coin has probability $p$ of landing heads, and $1-p$ of landing tails. If the coin is tossed independently $n$ times, then the probability of exactly $k$ heads is $\binom{n}{k}p^k(1-p)^{n-k}$.

In our case, $n=5$, $p=1/2$, and $k=1$.

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You're correct that there are $\;2^5 = 32$ possible outcomes of tossing 5 coin.

There are $\binom{5}{1} = 5$ of these outcomes which contain exactly one head. Indeed, these possible outcomes are precisely those listed below:

$(1)\quad H\;T\;T\;T\;T$
$(2)\quad T\;H\;T\;T\;T$
$(3)\quad T\;T\;H\;T\;T$
$(4)\quad T\;T\;T\;H\;T$
$(5)\quad T\;T\;T\;T\;H$

That gives us a probability of $\;\dfrac{5}{32}\;$ that exactly one head will face up upon tossing $5$ fair coins.

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Hint: How many ways are there of choosing which coin out of the five will come up "head"? How many possible outcomes are there total?

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