Two balls are selected at random one by one without replacement from a bag containing 4 white and 6 black balls. If the probability that the first selected ball is black, given that the second selected ball is also black, is $\frac mn$, where $\gcd(m,n)=1$, then $m+n$ is equal to :

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $P(B_1\cap B_2)=\frac6{10}\cdot\frac59=\frac13$ and $P(B_2)=\frac35$, so the conditional probability is $\frac{1/3}{3/5}=\frac59$ and $m+n=14$. Answer (2).

Q. Two balls are selected at random one by one without replacement from a bag containing 4 white and 6 black balls. If the probability that the first selected ball is black, given that the second selected ball is also black, is $\frac mn$, where $\gcd(m,n)=1$, then $m+n$ is equal to :

Explanation

Reviewed by Dr. Abubacker

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