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13 Feb 2020, 07:26
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
74% (02:30) correct
26%
(02:50)
wrong
based on 119
sessions
History
Date
Time
Result
Not Attempted Yet
A bag contains 4 red, 4 green and 4 white balls. Three balls are picked up one by one without replacement. What is the probability that there will be at least one red ball ?
A. 41/55
B. 3/64
C. 40/55
D. 41/50
E. 42/55
A. 41/55
B. 3/64
C. 40/55
D. 41/50
E. 42/55
13 Feb 2020, 09:11
Kudos
Bookmarks
We'd get several cases answering the question directly (exactly one red, exactly two red, exactly three red). So it will be easier to reverse the problem: if we can find the probability we get zero red balls, we can subtract that from 1 to answer the question. We have an 8/12 chance to avoid a red ball on the first selection, and then because we're selecting without replacement, we'll have a 7/11 chance on the next selection, and a 6/10 chance on the last selection. Multiplying, the probability we get zero red balls is
(8/12)(7/11)(6/10) = (2/5)(7/11) = 14/55
and subtracting from 1, the probability we get one or more red balls is 41/55.
(8/12)(7/11)(6/10) = (2/5)(7/11) = 14/55
and subtracting from 1, the probability we get one or more red balls is 41/55.
13 Feb 2020, 09:26
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Bunuel
We want P(select at least 1 red ball)
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 red ball) = 1 - P(getting NOT getting at least 1 red ball)
= 1 - P(getting ZERO red balls)
P(getting ZERO red balls)
P(getting ZERO red balls) = P(1st pick is non-red AND 2nd pick is non-red AND 3rd pick is non-red)
= P(1st pick is non-red) x P(2nd pick is non-red) x P(3rd pick is non-red)
= 8/12 x 7/11 x 6/10
= 14/55
So, P(getting at least 1 red ball) = 1 - 14/55
= 41/55
Answer: A
Cheers,
Brent
