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?
1/3
3/8
1/2
5/8
11/16
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We have 2 possibilities:
1-Mary picks a Withe ball and then John picks a Blue ball
3/8 * 5/7 = 15/56
2-Mary picks a Blue ball and then John picks a Blue ball
5/8 * 4/7 = 20/56
so (15+20)/56= 35/56 ==> 5/8 is the answer!
1-Mary picks a Withe ball and then John picks a Blue ball
3/8 * 5/7 = 15/56
2-Mary picks a Blue ball and then John picks a Blue ball
5/8 * 4/7 = 20/56
so (15+20)/56= 35/56 ==> 5/8 is the answer!
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No math is needed here if we understand the following concept:sachindia wrote: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?
1/3
3/8
1/2
5/8
11/16
The probability of selecting X on the NTH pick is equal to the probability of selecting X on the FIRST pick.
Thus, P(blue on the 2nd pick) = P(blue on the first pick) = 5/8.
The correct answer is D.
The answer would be the same even if the problem were as follows:
P(blue on the 3rd pick) = P(blue on the 1st pick) = 5/8.A basket contains 3 white and 5 blue balls. Mary will extract TWO BALLS at random and keep them. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?
Other problems that test the same concept:
https://www.beatthegmat.com/probablity-ques-t60161.html
https://www.beatthegmat.com/manhattan-pr ... 89481.html (2 posts)
https://www.beatthegmat.com/a-box-contai ... 51368.html
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These are two mutually exclusive events and occurrence of one is not dependent on occurrence of the other as the ball is placed back into bag after first pick up.
Since Mary placed the ball back again so when John will be picking up the ball he will have same number of balls in bag to choose from as there were initially i.e 5 blue and 3 white.
Hence probability of John picking up blue ball will be 5/8.
Since Mary placed the ball back again so when John will be picking up the ball he will have same number of balls in bag to choose from as there were initially i.e 5 blue and 3 white.
Hence probability of John picking up blue ball will be 5/8.
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hey , thanks a lot mitch for this. Does this work for problems with and without replacement?GMATGuruNY wrote:No math is needed here if we understand the following concept:sachindia wrote: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?
1/3
3/8
1/2
5/8
11/16
The probability of selecting X on the NTH pick is equal to the probability of selecting X on the FIRST pick.
Thus, P(blue on the 2nd pick) = P(blue on the first pick) = 5/8.
The correct answer is D.
The answer would be the same even if the problem were as follows:
P(blue on the 3rd pick) = P(blue on the 1st pick) = 5/8.A basket contains 3 white and 5 blue balls. Mary will extract TWO BALLS at random and keep them. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?
Other problems that test the same concept:
https://www.beatthegmat.com/probablity-ques-t60161.html
https://www.beatthegmat.com/manhattan-pr ... 89481.html (2 posts)
https://www.beatthegmat.com/a-box-contai ... 51368.html
Regards,
Sach
Sach
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GMATGuruNY
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Yes.sachindia wrote:hey , thanks a lot mitch for this. Does this work for problems with and without replacement?GMATGuruNY wrote:No math is needed here if we understand the following concept:sachindia wrote: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?
1/3
3/8
1/2
5/8
11/16
The probability of selecting X on the NTH pick is equal to the probability of selecting X on the FIRST pick.
Thus, P(blue on the 2nd pick) = P(blue on the first pick) = 5/8.
The correct answer is D.
The answer would be the same even if the problem were as follows:
P(blue on the 3rd pick) = P(blue on the 1st pick) = 5/8.A basket contains 3 white and 5 blue balls. Mary will extract TWO BALLS at random and keep them. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?
Other problems that test the same concept:
https://www.beatthegmat.com/probablity-ques-t60161.html
https://www.beatthegmat.com/manhattan-pr ... 89481.html (2 posts)
https://www.beatthegmat.com/a-box-contai ... 51368.html
If a bag contains 4 blue marbles and 5 white marbles, and 3 marbles are selected at random -- WITH OR WITHOUT REPLACEMENT -- the probability that the THIRD marble selected is blue is equal to the probability that the FIRST marble selected is blue: 4/9.
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For more information, please email me (Mitch Hunt) at [email protected].
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
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Brent@GMATPrepNow
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This question reminds me of my childhood, when my friends and I would sometimes "draw straws" to randomly select one person to do something (often either work, like getting wood for the fire, or dumb, like eating something that shouldn't be eaten).sachindia wrote: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?
1/3
3/8
1/2
5/8
11/16
So, someone would hold up n pieces of grass (for n guys), and one of those pieces was very short. The person who selected the shortest piece was the one who had to perform the task.
There was always one guy who wanted to choose his piece last. His reasoning was that his chances of drawing the shortest piece were minimized since every person before him had a chance of drawing the short piece before it got to his turn.
The truth of the matter is that each of the n guys had a 1/n chance of selecting the shortest piece, regardless of the order in which they selected.
The same applies to the original question here. 5 of the 8 balls are blue, so P(John selects a blue ball) = 5/8
Cheers,
Brent
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This is interesting. Where does this concept come from? I've not seen it used anywhere else. Thanks.GMATGuruNY wrote:No math is needed here if we understand the following concept:sachindia wrote: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?
1/3
3/8
1/2
5/8
11/16
The probability of selecting X on the NTH pick is equal to the probability of selecting X on the FIRST pick.
Thus, P(blue on the 2nd pick) = P(blue on the first pick) = 5/8.
The correct answer is D.
The answer would be the same even if the problem were as follows:
P(blue on the 3rd pick) = P(blue on the 1st pick) = 5/8.A basket contains 3 white and 5 blue balls. Mary will extract TWO BALLS at random and keep them. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?
Other problems that test the same concept:
https://www.beatthegmat.com/probablity-ques-t60161.html
https://www.beatthegmat.com/manhattan-pr ... 89481.html (2 posts)
https://www.beatthegmat.com/a-box-contai ... 51368.html
But Mary keeps the ball...How is it the same if she didn't keep it? Thank you.Brent@GMATPrepNow wrote:This question reminds me of my childhood, when my friends and I would sometimes "draw straws" to randomly select one person to do something (often either work, like getting wood for the fire, or dumb, like eating something that shouldn't be eaten).sachindia wrote: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?
1/3
3/8
1/2
5/8
11/16
So, someone would hold up n pieces of grass (for n guys), and one of those pieces was very short. The person who selected the shortest piece was the one who had to perform the task.
There was always one guy who wanted to choose his piece last. His reasoning was that his chances of drawing the shortest piece were minimized since every person before him had a chance of drawing the short piece before it got to his turn.
The truth of the matter is that each of the n guys had a 1/n chance of selecting the shortest piece, regardless of the order in which they selected.
The same applies to the original question here. 5 of the 8 balls are blue, so P(John selects a blue ball) = 5/8
Cheers,
Brent
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Let's solve this question a different way.sachindia wrote: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?
1/3
3/8
1/2
5/8
11/16
P(John gets blue ball) = P(Mary gets white ball and John gets blue ball OR Mary gets blue ball and John gets blue ball
= P(Mary gets white ball and John gets blue ball) + P(Mary gets blue ball and John gets blue ball)
= (3/8 x 5/7) + (5/8 x 4/7)
= 15/56 + 20/56
= 35/56
= 5/8
Answer: still D
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In an ordinary deck of cards, 1/4 of the cards are hearts. If a magician spreads out a deck of cards on a table, and asks you to pick one, and you pick the second card from the top, I think everyone would agree the probability that card is a heart is 1/4.
If instead, the magician picks the top card off the deck and keeps it for herself, then asks you to pick the second card from the deck, you're picking exactly the same card as in the situation above. The probability it is a heart must be the same as before - it must be 1/4.
That's the principle Mitch is using above. The probability the top card in a deck is a heart is the same as the probability the second card, or the eleventh card, or the 41st card is a heart. It doesn't matter if you reach into the deck and take a card from the middle, or if someone removes the earlier cards one-by-one before you pick your card.
The key here is that you have no information about the cards that have been removed, so they neither make it more nor less likely that subsequent cards are hearts. If you knew something about the removed cards ,then that would change everything - then you'd be able to use that knowledge to adjust the probabilities for subsequent card selections. But in the question in the OP, you know nothing about Mary's selection, so the probability John picks a blue marble is still 5/8.
If instead, the magician picks the top card off the deck and keeps it for herself, then asks you to pick the second card from the deck, you're picking exactly the same card as in the situation above. The probability it is a heart must be the same as before - it must be 1/4.
That's the principle Mitch is using above. The probability the top card in a deck is a heart is the same as the probability the second card, or the eleventh card, or the 41st card is a heart. It doesn't matter if you reach into the deck and take a card from the middle, or if someone removes the earlier cards one-by-one before you pick your card.
The key here is that you have no information about the cards that have been removed, so they neither make it more nor less likely that subsequent cards are hearts. If you knew something about the removed cards ,then that would change everything - then you'd be able to use that knowledge to adjust the probabilities for subsequent card selections. But in the question in the OP, you know nothing about Mary's selection, so the probability John picks a blue marble is still 5/8.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
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Another way to demonstrate this:
Suppose that there are x marbles that I want in a jar that contains y marbles in all. My friend gets to pick and keep a marble first, then it's my turn.
If my friend gets one of the nice marbles, then my chance of getting one on my pick is (x - 1)/(y - 1).
If my friend gets one of the crummy marbles, then my chance getting a nice on on my pick is x/(y - 1).
My friend's chance of getting a nice marble is x/y, and my friend's chance of getting a crummy marble is (y - x)/y, so my chance of getting a nice marble is
(x/y)*(x-1)/(y-1) + (y - x)/y * x/(y - 1) => x/y
Suppose that there are x marbles that I want in a jar that contains y marbles in all. My friend gets to pick and keep a marble first, then it's my turn.
If my friend gets one of the nice marbles, then my chance of getting one on my pick is (x - 1)/(y - 1).
If my friend gets one of the crummy marbles, then my chance getting a nice on on my pick is x/(y - 1).
My friend's chance of getting a nice marble is x/y, and my friend's chance of getting a crummy marble is (y - x)/y, so my chance of getting a nice marble is
(x/y)*(x-1)/(y-1) + (y - x)/y * x/(y - 1) => x/y
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As a fun aside, this problem is a nice primer on the perils of insider trading. Let's say that I'm a bookmaker and decide to take bets on this marble drawing. Prior to Mary's draw, John is 62.5% to draw blue, so the fair odds on that are 3 to 5: if you want to bet that John will draw blue, I'll wager 60¢ against every $1 that you're willing to risk.
After Mary's draw, however, *someone* -- Mary, at least (unless she doesn't look at the marble), and anyone she signals -- knows that the true odds have changed. If Mary drew white, then John's chances of blue are HIGHER, and if Mary drew blue, then John's chances of blue are lower. In either case, Mary now has better information than I do, and can outfox me in the betting! From John's perspective, or from any unknowing observer's, the best guess of his chances is still 62.5%, but anyone taking bets at this price can be defeated.
After Mary's draw, however, *someone* -- Mary, at least (unless she doesn't look at the marble), and anyone she signals -- knows that the true odds have changed. If Mary drew white, then John's chances of blue are HIGHER, and if Mary drew blue, then John's chances of blue are lower. In either case, Mary now has better information than I do, and can outfox me in the betting! From John's perspective, or from any unknowing observer's, the best guess of his chances is still 62.5%, but anyone taking bets at this price can be defeated.
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Scott@TargetTestPrep
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The probability that John will extract a blue ball is the probability that he will extract it after Mary extracts a blue ball plus the probability that he will extract it after Mary extracts a white ball. That is,
P(Mary extracts a blue ball followed by John a blue ball) + P(Mary extracts a white ball followed by John a blue ball)
5/8 * 4/7 + 3/8 * 5/7
20/56 + 15/56
35/56
5/8
Answer: D
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