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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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  • $\begingroup$ Check the answers here for the general approach. $\endgroup$ – Ayman Hourieh Jul 11 '13 at 0:17
  • $\begingroup$ @AymanHourieh: why that approach is general? I see the same question there. Do you say that the question there is different because it is exact duplicate? $\endgroup$ – Val Jul 11 '13 at 0:25
  • $\begingroup$ @Val The question is similar, but many of the answers give the general approach. $\endgroup$ – Ayman Hourieh Jul 11 '13 at 0:28
  • $\begingroup$ If questions like this are not assumed to be general then they must be closed as "too localized". If that question is not too localized, then this is not either and they are duplicate and this must be closed as duplicate. You will not stackoverflow this probelm for every combination of ${n\choose m}$. It makes no sense. Useful questions are supposed to be abstract and ask for a method rather than simple answer. $\endgroup$ – Val Jul 11 '13 at 0:43
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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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