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# M14-01

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Math Expert
Joined: 02 Sep 2009
Posts: 54496

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16 Sep 2014, 00:50
00:00

Difficulty:

45% (medium)

Question Stats:

73% (01:17) correct 27% (01:37) wrong based on 85 sessions

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A basket contains 3 white and 5 blue balls. Mary will extract one ball at random and keep it. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?

A. $$\frac{1}{3}$$
B. $$\frac{3}{8}$$
C. $$\frac{1}{2}$$
D. $$\frac{5}{8}$$
E. $$\frac{11}{16}$$

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Joined: 02 Sep 2009
Posts: 54496

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16 Sep 2014, 00:50
Official Solution:

A basket contains 3 white and 5 blue balls. Mary will extract one ball at random and keep it. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?

A. $$\frac{1}{3}$$
B. $$\frac{3}{8}$$
C. $$\frac{1}{2}$$
D. $$\frac{5}{8}$$
E. $$\frac{11}{16}$$

There are 2 different scenarios to evaluate:

1. Mary will extract a white ball, John will extract a blue ball. The probability of this is $$\frac{3}{8} * \frac{5}{7} = \frac{15}{56}$$.

2. Mary will extract a blue ball, John will extract a blue ball. The probability of this is $$\frac{5}{8} * \frac{4}{7} = \frac{20}{56}$$.

The overall probability that John will extract a blue ball = $$\frac{15}{56} + \frac{20}{56} = \frac{35}{56} = \frac{5}{8}$$.

Alternative Explanation

The initial probability of drawing blue ball is $$\frac{5}{8}$$. Without knowing the other results, the probability of drawing blue ball will not change for ANY successive drawing: second, third, fourth... There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).

If Mary extracts not 1 but 7 balls at random and keep them, then even after that, the probability that the last ball left for John is blue still would be $$\frac{5}{8}$$.

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Joined: 24 Dec 2015
Posts: 56
Location: India
Concentration: Marketing, Technology
GMAT 1: 710 Q50 V35
GPA: 3.8
WE: Engineering (Computer Software)

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04 Jul 2016, 19:59
I'm confused. Can someone help me here?

Probability = Favourable Outcomes/Total possible outcomes

In this case, since the first pick is not being replaced, shouldn't the Probability of the second pick be 5/7+4/7 ?
Math Expert
Joined: 02 Sep 2009
Posts: 54496

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05 Jul 2016, 01:47
winionhi wrote:
I'm confused. Can someone help me here?

Probability = Favourable Outcomes/Total possible outcomes

In this case, since the first pick is not being replaced, shouldn't the Probability of the second pick be 5/7+4/7 ?

The probability that John will extract a blue ball with second pick is 5/7 only if Mary extracts a white ball with her first pick.
The probability that John will extract a blue ball with second pick is 4/7 only if Mary extracts a blue ball with her first pick.

So, you should take the first pick into the account as shown in the solution above.
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Joined: 28 Feb 2016
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28 May 2017, 22:56
In the question, it should be made clear whether Mary keeps it back into the basket (replacement case) or keeps it with her (no replacement case). Just saying "keeps it" might create confusion.
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Joined: 02 Sep 2009
Posts: 54496

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28 May 2017, 23:29
akmi88 wrote:
In the question, it should be made clear whether Mary keeps it back into the basket (replacement case) or keeps it with her (no replacement case). Just saying "keeps it" might create confusion.

I don't think there is an ambiguity there. Extracts and keeps it means she does not put it back.
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Joined: 10 Nov 2014
Posts: 15
Location: United Kingdom
WE: Information Technology (Consulting)

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20 Aug 2017, 07:55
Bunuel wrote:
Official Solution:

A basket contains 3 white and 5 blue balls. Mary will extract one ball at random and keep it. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?

A. $$\frac{1}{3}$$
B. $$\frac{3}{8}$$
C. $$\frac{1}{2}$$
D. $$\frac{5}{8}$$
E. $$\frac{11}{16}$$

There are 2 different scenarios to evaluate:

1. Mary will extract a white ball, John will extract a blue ball. The probability of this is $$\frac{3}{8} * \frac{5}{7} = \frac{15}{56}$$.

2. Mary will extract a blue ball, John will extract a blue ball. The probability of this is $$\frac{5}{8} * \frac{4}{7} = \frac{20}{56}$$.

The overall probability that John will extract a blue ball = $$\frac{15}{56} + \frac{20}{56} = \frac{35}{56} = \frac{5}{8}$$.

Alternative Explanation

The initial probability of drawing blue ball is $$\frac{5}{8}$$. Without knowing the other results, the probability of drawing blue ball will not change for ANY successive drawing: second, third, fourth... There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).

If Mary extracts not 1 but 7 balls at random and keep them, then even after that, the probability that the last ball left for John is blue still would be $$\frac{5}{8}$$.

Hi Bunnuel,

Could you please elaborate a bit on the alternative solution? Why would 1st draw not change the probability of the second draw as the set from where the second draw is made would consist just of 7 balls and we aren't sure on the count of BLUE and WHITE in this draw as we do not know the result of the first.

Did it by the first process but took me close to 3 mins.

Best-
Amit
Math Expert
Joined: 02 Sep 2009
Posts: 54496

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21 Aug 2017, 02:47
1
Amit989 wrote:
Bunuel wrote:
Official Solution:

A basket contains 3 white and 5 blue balls. Mary will extract one ball at random and keep it. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?

A. $$\frac{1}{3}$$
B. $$\frac{3}{8}$$
C. $$\frac{1}{2}$$
D. $$\frac{5}{8}$$
E. $$\frac{11}{16}$$

There are 2 different scenarios to evaluate:

1. Mary will extract a white ball, John will extract a blue ball. The probability of this is $$\frac{3}{8} * \frac{5}{7} = \frac{15}{56}$$.

2. Mary will extract a blue ball, John will extract a blue ball. The probability of this is $$\frac{5}{8} * \frac{4}{7} = \frac{20}{56}$$.

The overall probability that John will extract a blue ball = $$\frac{15}{56} + \frac{20}{56} = \frac{35}{56} = \frac{5}{8}$$.

Alternative Explanation

The initial probability of drawing blue ball is $$\frac{5}{8}$$. Without knowing the other results, the probability of drawing blue ball will not change for ANY successive drawing: second, third, fourth... There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).

If Mary extracts not 1 but 7 balls at random and keep them, then even after that, the probability that the last ball left for John is blue still would be $$\frac{5}{8}$$.

Hi Bunnuel,

Could you please elaborate a bit on the alternative solution? Why would 1st draw not change the probability of the second draw as the set from where the second draw is made would consist just of 7 balls and we aren't sure on the count of BLUE and WHITE in this draw as we do not know the result of the first.

Did it by the first process but took me close to 3 mins.

Best-
Amit

Because we don't know the result of the first draw.

Consider this imagine 8 balls in a row. Now, what is the probability that blue ball is 1st in that row? 5/8. What is the probability that it's 2nd? Again 5/8. What is the probability that he's 8th? 5/8.

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Hope it helps.
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Re: M14-01   [#permalink] 21 Aug 2017, 02:47
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# M14-01

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