A box contains ten marbles, seven of which are black and three are red. Three...

A box contains ten marbles, seven of which are black and three are red. Three marbles are drawn one after the other without replacement. Find the probability of choosing at least two black marbles.?

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Answers (4)

Jace
1 week ago
P(X ≥ 2) = 1 - P(X = 0) - P(X = 1)
= 1 -(7/0)*(3/3)/(10/3)-(7/1)*(3/2)/(10/3)
= 49/60 or 0.8167
Jr🕴
1 week ago
P(X ≥ 2) = 1 - P(X = 0) - P(X = 1)
= 1 -(7/0)*(3/3)/(10/3)-(7/1)*(3/2)/(10/3)
= 49/60 or 0.8167 ans
isaaq
1 week ago
Ways (permutations) of randomly drawing 3 marbles from among 10 randomly distributed marbles = 10!/(7!) = 720.
Ways of choosing in the order as given: 1 R marble =3, 1 B marble = 7, 2nd R marble = 2. Ways of choosing marbles in accordance with the question = 3*7*2 = 42. Probability of doing so = 42/720 = 7/120 = .05833333…~ 5.8%.
The same result is obtained by multiplying probabilities as follows:
(3/10)*(7/9)*(2/8) = 42/720 = 7/120 = .05833333…~ 5.83%.
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