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Question
A bag contains 5 red marbles and 3 black marbles. Three marbles are drawn one by one without replacement. What is the probability that at least one of the three marbles drawn be black, if the first marble is red?
Solution
Let red marbles be represented with R and black marble with B.
The following three conditions are possible
If atleast one of the three marbles drawn be black and the first marble is red.
(i) E1: II ball is black and III is red
(ii) E2: II ball is black and III is also black
(iii) E3: II ball is red and III is black
∴ P(E1) = `"P"("R"_1) . "P"("B"_1/"R"_1) . "P"("R"_2/("R"_1"B"_1))`
= `5/8 * 3/7 * 4/6`
= `60/336`
= `5/28`
P(E2) = `"P"("R"_1) . "P"("B"_1/"R"_1) . "P"("B"_2/("R"_1"B"_1))`
= `5/8 * 3/7 * 2/6`
= `30/336`
= `5/36`
And P(E3) = `"P"("R"_1) . "P"("R"_2/"R"_1) . "B"("B"_1/("R"_1"R"_2))`
= `5/8 * 4/7 * 3/6`
= `60/336`
= `5/28`
∴ P(E) = P(E1) + P(E2) + P(E3) = `5/28 + 5/56 + 5/28 = 25/56`
Hence the required probability is `25/56`.
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