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Question
A bag contains 3 red marbles and 4 blue marbles. Two marbles are drawn at random without replacement. If the first marble drawn is red, what is the probability the second marble is blue?
Solution
Let A ≡ the event that first marble is red
B ≡ the event that second marble is blue
∴ the required probability = P(A ∩ B)
= P(A) = `"P"("B"//"A")` ...(1)
Clearly P(A) = `3/(3 + 4) = 3/7`
If first marble is not replaced, bag will contain 2 red marbles and 4 blue marbles, i.e., altogether 6 marbles when second marble is drawn.
Probability that second marble is blue under the condition that first red marble is not replaced
= `"P"("B"//"A") = 4/6 = 2/3`
∴ from (1), we get,
P(A ∩ B) = `3/7 xx 2/3 = 2/7`.
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