It is "(m+f)!/2!(m+f-2)!" if there are no replacements. That is, once you select a student (male/female), you need to exclude that selected student from the total.

Here, replacement is considered.

Probability of selecting a student, for the first pick = (m+f)
Probability of selecting a student, for the second pick = (m+f) (since, the total doesn't change -- student replaced)
Probability of selecting two = (m+f)^2

Yup the answer to this question is indeed B
Clearly we cannot find the probability using statement A
The statement B is sufficient

ANSWER: B

Can someone help me clear my doubt pls .

Suppose we have 10 red and 4 blue balls . What is the probability of picking 1 red and 1 blue ball with replacement ?

My methodology - 10/14 * 4/13 . Since it is with replacement the total(denominator) itself changes . Why isn't this the case with the solution provided for B ? How does the total remain constant for both m/(m+f) and f / (m+f) .

Where am I going wrong?

Deepthi Subbu wrote:Can someone help me clear my doubt pls .

Suppose we have 10 red and 4 blue balls . What is the probability of picking 1 red and 1 blue ball without replacement ?

My methodology - 10/14 * 4/13 . Since it is with replacement the total(denominator) itself changes . Why isn't this the case with the solution provided for B ? How does the total remain constant for both m/(m+f) and f / (m+f) .

Where am I going wrong?

The stem reads "without replacement" but you reason "Since it is with replacement". Furthermore, when two balls are drawn with replacement, it's understood that they are drawn one by one with replacement and hence the denominator '14' won't change in your calculation.

vkb001 wrote:It is "(m+f)!/2!(m+f-2)!" if there are no replacements. That is, once you select a student (male/female), you need to exclude that selected student from the total.

Here, replacement is considered.

Probability of selecting a student, for the first pick = (m+f)
Probability of selecting a student, for the second pick = (m+f) (since, the total doesn't change -- student replaced)
Probability of selecting two = (m+f)^2

read "number of ways" in place of Probability here

amitakalra88 wrote:I can't understand how students can be chosen WITH REPLACEMENT. As in, if the same student is selected twice then the two member committee wont be formed at all, will it?

Also, why are 1M,1F and 1F,1M being treated as separate cases? I cant understand how this is related to replacement. We are taking only combinations here, not permutations, so I don't understand the explanation.

Please help.
Thanks in advance.

The given question is not bothered about forming a two member committee at all. It's simply suggesting us that the selection of one student is made two times from the same lot of m + f students. Had it been made for say 3 or 4 number of times, the author would still have been technically free to write it as

"If three or four students are chosen at random with replacement from a certain class..."

Since the operation of selection of one student is done twice from the same total, 1 m and 1 f must have to be taken as either 1m 1f or 1f 1m. Remember that the two are not taken in a single go, hence two students of opposite gender can be obtained in either of the two ways

1. Having a male in the first pick and a female in the second pick, or

2. Having a female in the first pick and a male in the second pick.

sanju09 wrote:

Deepthi Subbu wrote:Can someone help me clear my doubt pls .

Suppose we have 10 red and 4 blue balls . What is the probability of picking 1 red and 1 blue ball without replacement ?

My methodology - 10/14 * 4/13 . Since it is with replacement the total(denominator) itself changes . Why isn't this the case with the solution provided for B ? How does the total remain constant for both m/(m+f) and f / (m+f) .

Where am I going wrong?

The stem reads "without replacement" but you reason "Since it is with replacement". Furthermore, when two balls are drawn with replacement, it's understood that they are drawn one by one with replacement and hence the denominator '14' won't change in your calculation.

Sorry for the confusion.I was supposed to mean with replacement . Can you explain in that case?

Deepthi Subbu wrote:

sanju09 wrote:

Deepthi Subbu wrote:Can someone help me clear my doubt pls .

Suppose we have 10 red and 4 blue balls . What is the probability of picking 1 red and 1 blue ball without replacement ?

My methodology - 10/14 * 4/13 . Since it is with replacement the total(denominator) itself changes . Why isn't this the case with the solution provided for B ? How does the total remain constant for both m/(m+f) and f / (m+f) .

Where am I going wrong?

The stem reads "without replacement" but you reason "Since it is with replacement". Furthermore, when two balls are drawn with replacement, it's understood that they are drawn one by one with replacement and hence the denominator '14' won't change in your calculation.

Sorry for the confusion.I was supposed to mean with replacement . Can you explain in that case?

The probability of picking 1 red and 1 blue ball with replacement, in the case cited by you

= probability that first draw is a red × probability that second draw is a blue + probability that first draw is a blue × probability that second draw is a red (from the same set of 14)

= 10/14 × 4/14 + 4/14 × 10/14

= 80/196

= [spoiler]20/49[/spoiler]

Thanks sanju09 . Can you explain the same w/o replacement ?

Deepthi Subbu wrote:Thanks sanju09 . Can you explain the same w/o replacement ?

The probability of picking 1 red and 1 blue ball without replacement

= probability that first draw is a red X probability that second draw is a blue + probability that first draw is a blue X probability that second draw is a red

= 10/14 X 4/13 + 4/14 X 10/13

= 80/182

= [spoiler]40/91[/spoiler]

i am still not able to get this question but tomada explained it in a much better way.if the probability of picking(1M and 1F) IS 21/50 then for 2 males and 2females it will be 42/50.is there any other way to understand this question except the ones explained earlier.

So, is the answer B?

How do you arrive at 1-21/50 conceptually? and why does it equal 29/50? Can you explain thoroughly your reasoning with statement two?

Ok, Answer is B.

From stem:

Required Prob= (m^2+f^2)/(m+f)^2

Statement 1: Does not tell us anything conclusive.

Statement 2: (mf)/(m+f)^2 = 21/50

From question stem= m^2+f^2+2mf-2mf / (m+f)^2
=> 1- 2X (mf)/(m+f)^2
=> 1- 2X 21/50
=> 1- 42/50 => 8/50
=> 4/25

Hope that helps.

clearly statement 1 is not sufficient

for 2 . probablity of 1m & 1f is 21/50 = 7*3/5*5*2

=> 7/10 * 3/5 thus by this we can find the probablity

2 is sufficient