# 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? - Mathematics and Statistics

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?

#### 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
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