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Replacement in probability : concept

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by bblast » Sun Jul 24, 2011 12:07 am
1>What is the probability that one of the two integers randomly selected from range 20-29 is prime and the other is a multiple of 3?

[spoiler]{2/10 * 3/10} *2 = .12[/spoiler]


2>A basket contains 13 fruits: 5 apples and 8 pears. What is the probability that 2 fruits drawn from the basket at random will be pears?

[spoiler]8/13*7/12 = 14/39[/spoiler]




Both question above are from gmat club tests. My query is when do we consider that we are talking about replacements ? I mean the answer trap to the second question could have been 8/13*8/13 (if we considered that the fruit is replaced after withdrawal)


In my first question if we considered that the prime is not being put back then the solution would be {2/10 * 3/9} *2
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Source: — Problem Solving |

by knight247 » Sun Jul 24, 2011 1:02 am
We consider replacement only and only when it is stated or implicitly understood. In the first question we consider it as replacement simply because you picking a number from there does not deplete one number from the pool. But on the other hand if you had the numbers 20-29 written on chits of paper and placed in a bowl and 2 chits were picked out without replacement that would deplete numbers from the bowl. And in the second scenario the answer would be {2/10 * 3/9} *2. This concept is also applied to question 2


In question 2. Thinking outside the scope of mathematics....When a fruit is taken out of a basket the same fruit cannot be picked out of the basket again. You can only select from one of the remaining fruits. So taking a fruit out of the basket reduces the number of possible options by one.
Last edited by knight247 on Sun Jul 24, 2011 3:23 am, edited 1 time in total.
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by naveen451 » Sun Jul 24, 2011 2:37 am
in first question
{2/10*3/10}*2 is the ans

whatis the need of *2
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by knight247 » Sun Jul 24, 2011 3:08 am
Specific order is not mentioned in the problem. It could be P(Prime)*P(# divisible by3)+P(# divisible by3)*P(Prime). Hence the 2
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