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Here is the question :If we are told that among the 10 coins, at least three have landed on heads. What is the probability that there are at most five heads?

In other words,

P( at most five head | at least three have landed).

The solution says, $P( X\leq 5 | X\geq 3) = P( 3\leq X\leq 5)/P( X\geq3) \approx 0.601$

Which completely makes sense. But initially, I got this question wrong because I assumed that

$P(\text{at most five head} | \text{at least three have landed})$ is equivalent to $P(\text{at most 2 heads out of 7 toss}) = \binom{7}{2}(\frac{1}{2})^2(\frac{1}{2})^5\approx 0.1641$

Why can't I make this assumption? "Given that three have already landed head," don't I only care about $2$ more heads out of $7$ tosses? and aren't tosses all independent events?

I understand the correct method but I cannot resolve as to why my original solution is incorrect.

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  • $\begingroup$ Nice question. You have now got me confused as well :) $\endgroup$ – TMM May 1 '12 at 20:54
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Your calculation of $P(\text{at most $2$ heads out of $7$ tosses})$ is wrong because you must include $X=0$ and $X=1$ along with $X=2$. The correct result is $\approx 0.227$.

But that is not the main point. You know that three of the ten tosses are heads, but you don't know which ones. For instance, if you are given that the first three tosses landed on heads, then your result is OK. Does this make sense?

Edit: I should explain myself better. Suppose that you have tossed two coins. The probability of two heads, given that the first one is a head, is $1/2.$ The probability of two heads, given that at least one of them is a head, is $1/3.$ (Agreed?) This simple variant contains the main point of this apparent conflict.

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