SOLUTION: a box contains 10 marbles 7 of which are black and 3 are red . if 3 marbles is drawn at random without replacement find the probability of choosing (a) ome red one black and one r

Question 1174299: a box contains 10 marbles 7 of which are black and 3 are red . if 3 marbles is drawn at random without replacement find the probability of choosing (a) ome red one black and one red in that order (b) at least two black marbles (c) at most two black marbles

Found 2 solutions by ewatrrr, ikleyn:
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!

Hi 10 marbles: 7B and 3R if 3 marbles is drawn at random without replacement: find the probability of choosing (a) one red, one black and one red in that order: (3/10)(7/9)(2/8) (b) at least two black marbles 1 - No black marbles = 1-binomcdf(3, .7, 0) = 1 - .027 = .973 (c) at most two black marbles binomcdf(3, .7, 2) = .657 Wish You the Best in your Studies.

Answer by ikleyn(52909)   (Show Source): You can put this solution on YOUR website!
.
A box contains 10 marbles, 7 of which are black and 3 are red.
3 marbles is drawn at random without replacement. Find the probability of choosing
(a) one red one black and one red in that order
(b) at least two black marbles
(c) at most two black marbles
~~~~~~~~~~~~~~~

            The solution by @ewatrrr to parts (b) and (c) are absolutely WRONG.

            She mistakenly treats the probabilities in this problem as BINOMIAL DISTRIBUTION PROBLEMS, while THEY ARE NOT.

            So, I came to bring the correct solution/solutions.

(a) Red-Black-Red in this order P = = = . ANSWER (b) at least two black marbles P = P(3 black marbles) + P(2 black marbles and 1 red marble). P(3 black marbles) = = = = = ; P(2 black marbles and 1 red marble) = P(BBR) + P(BRB) + P(RBB) = = = = = . THEREFORE, P = + = = = . ANSWER (c) at most two black marbles P = 1 - P(3 black marbles) = 1 - = 1 - = 1 - = 1 - = . ANSWER

Solved.

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