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you flip three coins what is the probability that you get exactly 3 heads , given that you get at least 1 head ?
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P(A|B) = P(AB)/P(B) A = You get exactly 3 heads. B = You get at least 1 head.
\[\begin{array}{c|cccc} x & 0 & 1 & 2 & 3 \\\hline P(X=x) & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} \end{array}\]
so its 1/8 ?
no
use mtbender74 formula with the distribution I gave you
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noting that AB in this context means A and B.
Also noting that P(B) is P(X=1) + P(X=2) + P(X=3)...
1/7 :)
yes
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