A box contains 5 blue and 4 red balls. Two balls are drawn successively from the
box without replacement, and it is noted that the second one is red. What is the
probability that the first is also red?
(My answer is: 4/9 * 1 .is it true?)
[Math] A box contains 5 blue and 4 red balls. (probability problem)
combinatoricsprobabilityprobability theory
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Best Answer
So there are two things that can happen here. On the second pick, you are picking a red ball for sure. So we can have two different probabilities here. Why? Suppose the first ball you picked was blue. Then that blue / red combination probability would be $$\frac{5\cdot{4}}{9\cdot{8}}$$ Now suppose the first ball you picked was red. Then that red / red combination probability would be $$\frac{4\cdot{3}}{9\cdot{8}}$$ So let A be the event that you pick a red ball first and let B be the event that you pick a red ball second. Then we have the conditional probability: $$P(A|B)=\frac{P(A\cap{B})}{P(B)}=\frac{\frac{4\cdot{3}}{9\cdot{8}}}{\frac{5\cdot{4}+4\cdot{3}}{9\cdot{8}}}=\frac{12}{32}=\frac{3}{8}$$