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02 Aug 2021, 15:50
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A box with 5 white balls, and three blue balls from which four balls were selected. calculate:
Determine:
a. the probability of receiving two white balls.
b. the probability of receiving at most a white ball.
c. the probability of receiving at least two blue balls.
Determine:
a. the probability of receiving two white balls.
b. the probability of receiving at most a white ball.
c. the probability of receiving at least two blue balls.
bansalgaurav
Joined: 26 Mar 2021
Last visit: 20 Nov 2021
Posts: 112
Given Kudos: 36
Location: India
Concentration: General Management, Strategy
Schools: IIMB EPGP'23
GMAT 1: 640 Q50 V25
03 Aug 2021, 23:39
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Assuming No replacement and All picked at once, not in sequence.
white - 5 balls , blue - 3 balls ==> total = 8
Total number of cases(possible Combinations) of selecting 4 balls at once = 8C4
A: the probability of receiving two white balls = P(2 white and 2 blue)
Total number of combinations of 2 are white and 2 are blue = 5C2 x 3C2
Required Probab = 5C2 x 3C2 / 8C4
B : the probability of receiving at most a white ball = P(0 white) + P(1 white)
Now P(0 white) = P(all 4 blue ) which is not possible. Since 4 balls are drawn and the urn has only 3 blue balls.
the probability of receiving at most a white ball = P(1 white 3 Blue)
number of combinations of 1 white and 3 blue = 5C1 x 3C3
Required Probab = 5C1 x 3C3 / 8C4
c. the probability of receiving at least two blue balls. = P(2 blue) + P(3 blue)
Calculating P(2 Blue)
number of combinations of 2 blue and 2 white = 5C2 x 3C2
Required Probab = 5C2 x 3C2 / 8C4
Calculating P(3 blue)
number of combinations of 3 blue and 1 white = 5C1 x 3C3
Required Probab = 5C1 x 3C3 / 8C4
required Probab = 5C2 x 3C2 / 8C4 + 5C1 x 3C3 / 8C4
white - 5 balls , blue - 3 balls ==> total = 8
Total number of cases(possible Combinations) of selecting 4 balls at once = 8C4
A: the probability of receiving two white balls = P(2 white and 2 blue)
Total number of combinations of 2 are white and 2 are blue = 5C2 x 3C2
Required Probab = 5C2 x 3C2 / 8C4
B : the probability of receiving at most a white ball = P(0 white) + P(1 white)
Now P(0 white) = P(all 4 blue ) which is not possible. Since 4 balls are drawn and the urn has only 3 blue balls.
the probability of receiving at most a white ball = P(1 white 3 Blue)
number of combinations of 1 white and 3 blue = 5C1 x 3C3
Required Probab = 5C1 x 3C3 / 8C4
c. the probability of receiving at least two blue balls. = P(2 blue) + P(3 blue)
Calculating P(2 Blue)
number of combinations of 2 blue and 2 white = 5C2 x 3C2
Required Probab = 5C2 x 3C2 / 8C4
Calculating P(3 blue)
number of combinations of 3 blue and 1 white = 5C1 x 3C3
Required Probab = 5C1 x 3C3 / 8C4
required Probab = 5C2 x 3C2 / 8C4 + 5C1 x 3C3 / 8C4
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