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Edward has four pearls – two white and two black – which he can distri
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22 Oct 2021, 06:56
22 Oct 2021, 06:56
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Edward has four pearls – two white and two black – which he can distribute as he chooses between two identical bags. He must then choose a bag at random and pick one pearl at random from the bag he chose. If Edward distributes the pearls so as to maximize his chances of pulling a black pearl, what is the probability that Edward will pick a black pearl?
A. 1/2
B. 3/5
C. 2/3
D. 3/4
E. 1
A. 1/2
B. 3/5
C. 2/3
D. 3/4
E. 1
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Re: Edward has four pearls – two white and two black – which he can distri
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22 Oct 2021, 07:15
22 Oct 2021, 07:15
Carcass wrote:
Edward has four pearls – two white and two black – which he can distribute as he chooses between two identical bags. He must then choose a bag at random and pick one pearl at random from the bag he chose. If Edward distributes the pearls so as to maximize his chances of pulling a black pearl, what is the probability that Edward will pick a black pearl?
A. 1/2
B. 3/5
C. 2/3
D. 3/4
E. 1
A. 1/2
B. 3/5
C. 2/3
D. 3/4
E. 1
There aren't many different ways to distribute the balls, so let's list them:
1) Bag X: 0 marbles | Bag Y: BBWW
2) Bag X: B | Bag Y: BWW
3) Bag X: W | Bag Y: BBW
4) Bag X: BB | Bag Y: WW
5) Bag X: BW | Bag Y: BW
NOTE: Since the bags are considered identical, we can ignore equivalent distributions like Bag X: BWW | Bag Y: B, which is the same as the 2nd distribution in the list.
Some students may be able to automatically eliminate some answer choices upon examination.
Let's examine each case and determine P(Edward selects a black ball)
1) Bag X: 0 marbles | Bag Y: BBWW
P(black ball) = P(select bag Y AND select one of the black balls in that bag)
= P(select bag Y) x P(select one of the black balls in that bag)
= 1/2 x 2/4
= 1/4
2) Bag X: B | Bag Y: BWW
P(black ball) = P(select bag X AND select black ball in that bag OR select bag Y AND select black ball in that bag]
= [P(select bag X) x P(select black ball in that bag)] + [P(select bag Y) x P(select black ball in that bag)]
= [1/2 x 1] + [1/2 x 1/3]
= 1/2 + 1/6
= 4/6
= 2/3
3) Bag X: W | Bag Y: BBW
P(black ball) = P(select bag Y AND select one of the black balls in that bag)
= P(select bag Y) x P(select one of the black balls in that bag)
= 1/2 x 2/3
= 1/3
4) Bag X: BB | Bag Y: WW
P(black ball) = P(select bag X AND select one of the black balls in that bag)
= P(select bag X) x P(select one of the black balls in that bag)
= 1/2 x 1
= 1/2
5) Bag X: BW | Bag Y: BW
P(black ball) = P(select bag X AND select black ball in that bag OR select bag Y AND select black ball in that bag)]
= [P(select bag X) x P(select black ball in that bag)] + [P(select bag Y) x P(select black ball in that bag]
= [1/2 x 1/2] + [1/2 x 1/2]
= 1/4 + 1/4
= 1/2
Of all possible distributions, the highest probability is 2/3
Answer: C
Cheers,
Brent
gmatclubot
Re: Edward has four pearls – two white and two black – which he can distri [#permalink]
22 Oct 2021, 07:15
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