Sum

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?

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#### 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`.

Concept: Multiplication Theorem on Probability

Is there an error in this question or solution?

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