M14-01

Question

In a basket, there are 3 white balls and 5 blue balls. If Mary draws one ball at random and keeps it, what is the probability that John will subsequently draw a blue ball?

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

Bunuel · Accepted Answer

Official Solution:

In a basket, there are 3 white balls and 5 blue balls. If Mary draws one ball at random and keeps it, what is the probability that John will subsequently draw 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 two different scenarios to evaluate:
 
1. Mary extracts a white ball, and John extracts a blue ball. The probability of this occurring is  \frac{3}{8} * \frac{5}{7} = \frac{15}{56} .
 
2. Mary extracts a blue ball, and John extracts another blue ball. The probability of this happening is  \frac{5}{8} * \frac{4}{7} = \frac{20}{56} .
 
The overall probability that John will extract a blue ball is  \frac{15}{56} + \frac{20}{56} = \frac{35}{56}=\frac{5}{8} .
 
 Alternative Explanation 
 
The initial probability of drawing a blue ball is  \frac{5}{8} . Without knowing the outcomes of previous drawings, the probability of drawing a blue ball stays the same for any successive drawing: second, third, fourth, and so on. There's simply no reason to assume that one drawing is different from another when we're unaware of the outcomes of previous drawings.
 
If Mary were to extract not just 1 but 7 balls randomly and retain them, even then, the probability that the last ball left for John is blue would still be  \frac{5}{8} .

Answer: D

winionhi · Answer

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 ?

Bunuel · Answer

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.

Amit989 · Answer

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

Bunuel · Answer

The outcome of the first draw remains unknown to us. To illustrate, consider 8 balls lined up in a sequence. What's the likelihood that the first ball in that sequence is blue? It's 5/8. How about the second ball? It's still 5/8. And the probability for the eighth ball being blue? Again, 5/8. Similar questions to practice: https://gmatclub.com/forum/a-certain-te ... 31321.html https://gmatclub.com/forum/a-certain-te ... 27423.html https://gmatclub.com/forum/a-medical-re ... 27396.html https://gmatclub.com/forum/a-certain-cl ... 27730.html https://gmatclub.com/forum/a-certain-cl ... 91-20.html Hope it helps.

simonnguyen · Answer

You can rethink the question like the one below: 
How many ways to arrange 3 WHITE marbles & 5 BLUE marbles in a row with the 2nd marble is always blue ?

And here is a solution:
(1) Possible arrangement of 8 marbles in a row with 2nd marble is blue = 7! / (3!*4!)

Next step to calculate the probability:
(2) Possible arrangement of 8 marbles without any restriction = 8! / (3!*5*)

Probability --> (1) / (2) =  5/8

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

MalachiKeti · Answer

Bunuel 
Why cant the first case be (5c1*3c1)/8c2

Bunuel · Answer

That's because we need WB in this specific order, with the first ball being white and the second ball being blue. However, the denominator, 8C2, gives the pairs without considering order, so we need to multiply the denominator by 2 to account for the order.

MegB07 · Answer

Hi Bunuel,

I understood your approach but not sure how my approach is wrong.

So mary draws one ball and john draws a blue ball= 1/8x5/7 = 5/56
Mary draws one blue ball and john draws a blue ball = 1/8x4/7 = 4/56

Where am i going wrong with the approach?

Bunuel · Answer

You used 1/8 instead of 3/8 and 1/8 instead of 5/8. It should be 3/8 * 5/7 + 5/8 * 4/7 = 35/56 = 5/8.

pdave · Answer

HI  Bunuel , in this case why did we add the probabilities and not multiply?

Bunuel · Answer

We add because John getting a blue ball can happen in  two separate, non-overlapping cases  based on what Mary drew. Since those cases can’t both occur, you combine them by  adding  their probabilities.

M14-01

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