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M14-01

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M14-01 [#permalink]

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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. \(\frac{11}{16}\)
[Reveal] Spoiler: OA

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Re M14-01 [#permalink]

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New post 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 \(\frac{5}{8}\).


Answer: D
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Re: M14-01 [#permalink]

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New post 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 ?

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Re: M14-01 [#permalink]

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New post 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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Re: M14-01 [#permalink]

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New post 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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New post 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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Re: M14-01 [#permalink]

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New post 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 \(\frac{5}{8}\).


Answer: D




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

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Re: M14-01 [#permalink]

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New post 21 Aug 2017, 02:47
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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 \(\frac{5}{8}\).


Answer: D




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.

Similar questions to practice:
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Hope it helps.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M14-01   [#permalink] 21 Aug 2017, 02:47
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