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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.
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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
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
