Three coins are flipped. If at least one of them comes up heads, then what is the probability that they all come up heads?
The answer to this is 1/7, as it is the #(ways to have all heads)/#(ways to have at least one head).
If the question had asked, probability of all heads given that the first flip is heads, then the answer would have been 1/4 (because 1/2 * 1/2).
Why are these two answers different? Why is it that the "at least one" is a weaker condition, when at least simply means that one of the three coins is heads, which is the same as choosing a specific head, like the first coin? I would greatly appreciate some intuition here. Thanks!
Why is the at least one case less likely? If anything, shouldn't it be more likely because it doesn't restrict a specific coin to come up heads?
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$\begingroup$General result: Let $A, B, C$ be events in a probability space, all with positive probability. If $A\subseteq B\subseteq C$, then:
$P(A \mid B) \ge P(A\mid C)$.
This follows because $A$ 'takes up more of ' $B$ than it does of $C$.
If you apply the above general result to these events in the space of flipping three coins:
$A$ is the event of all three coming up heads;
$B$ is the event of the first coin coming up heads; and
$C$ is the event of at least one coin coming up heads;
you get that $A$ given $B$ is more likely than $A$ given $C$, as desired.
$\endgroup$ 1 $\begingroup$$$ \begin{array}{c} & HHH \\ & HHt \\ & HtH \\ & tHH \\ & Htt \\ & tHt \\ & ttH \end{array} $$The seven outcomes in which at least one "head" appears are listed above. In just one instance, all three are "heads."
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