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A bag contains equal numbers of red, green, and yellow marbles. If Gee
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01 Nov 2021, 06:39
01 Nov 2021, 06:39
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A bag contains equal numbers of red, green, and yellow marbles. If Geeta pulls three marbles out of the bag, replacing each marble after she picks it, what is the probability that at least one will be red?
A. 8/27
B. 19/27
C. 20/27
D. 7/9
E. 22/27
A. 8/27
B. 19/27
C. 20/27
D. 7/9
E. 22/27
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Re: A bag contains equal numbers of red, green, and yellow marbles. If Gee
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01 Nov 2021, 06:49
01 Nov 2021, 06:49
Carcass wrote:
A bag contains equal numbers of red, green, and yellow marbles. If Geeta pulls three marbles out of the bag, replacing each marble after she picks it, what is the probability that at least one will be red?
A. 8/27
B. 19/27
C. 20/27
D. 7/9
E. 22/27
A. 8/27
B. 19/27
C. 20/27
D. 7/9
E. 22/27
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 marble) = 1 - P(not getting at least 1 red marble)
In other words, P(getting at least 1 red marble) = 1 - P(not getting ZERO red marbles)
P(getting ZERO red marbles)
P(getting ZERO red marbles) = P(1st marble is not red AND 2nd marble is not red AND 3rd marble is not red)
= P(1st marble is not red) x P(2nd marble is not red) x P(3rd marble is not red)
= 2/3 x 2/3 x 2/3
= 8/27
Aside: Since we have an equal number of each colored marble, P(red) = P(green) = P(yellow) = 1/3. So, p(not red) = 2/3 each time
So, P(getting at least 1 red marble) = 1 - 8/27
= 27/27 - 8/27
= 19/27
Answer: B
gmatclubot
Re: A bag contains equal numbers of red, green, and yellow marbles. If Gee [#permalink]
01 Nov 2021, 06:49
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