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ariellelouise
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ariellelouise
A bag contains 3 yellow and 5 blue equally sized marbles. Two marbles are randomly removed from the bag. What is the probability that 3 yellow and 3 blue marbles remain in the bag?

(A) 1/4
(B) 5/14
(C) 1/2
(D) 5/7
(E) 5/8

Given: A bag contains 3 yellow and 5 blue equally sized marbles. Two marbles are randomly removed from the bag.
Asked: What is the probability that 3 yellow and 3 blue marbles remain in the bag?

Total cases of selecting 2 marbles = 8C2 = 28
If 3 yellow and 3 blue marbles remain in the bag, then 2 blue marbles are taken out of 5 blue marbles
Number of favourable cases = 5C2 = 10

Probability = 10/28 = 5/14

IMO B
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ariellelouise
A bag contains 3 yellow and 5 blue equally sized marbles. Two marbles are randomly removed from the bag. What is the probability that 3 yellow and 3 blue marbles remain in the bag?

(A) 1/4
(B) 5/14
(C) 1/2
(D) 5/7
(E) 5/8

total marbles = 8
and since 3 yellow and 3 blue marble are to remain so only 2 blue are to be removed
we have
5/8* 4/7 ; 5/14
IMO B
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ariellelouise
A bag contains 3 yellow and 5 blue equally sized marbles. Two marbles are randomly removed from the bag. What is the probability that 3 yellow and 3 blue marbles remain in the bag?

(A) 1/4
(B) 5/14
(C) 1/2
(D) 5/7
(E) 5/8

P(3 yellow and 3 blue marbles REMAIN) = P(2 blue marbles are REMOVED)
= P(1st marble is blue AND 2nd marble is blue)
= P(1st marble is blue) x P(2nd marble is blue)
= 5/8 x 4/7
= 20/56
= 5/14

Answer: B

Cheers,
Brent
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