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PROBABILITY DS

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by cramya » Sat Nov 15, 2008 2:03 pm
If two students are chosen at random with replacement from a certain class, what is the probability that two male students or two female students are selected?

1) There are 41 male students in the class.
2) The probability of selecting one male and one female student is 15/41.

[spoiler]OA: B[/spoiler]
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Re: PROBABILITY DS

by logitech » Sat Nov 15, 2008 2:44 pm
cramya wrote:If two students are chosen at random with replacement from a certain class, what is the probability that two male students or two female students are selected?

1) There are 41 male students in the class.
2) The probability of selecting one male and one female student is 15/41.

[spoiler]OA: B[/spoiler]
Cramya in da house!!

Okay let me try this:

M=Males
F=Females

What we need is:

P(2males) + P(2females)

With replacement means there will always be M+F students ?

=[M/(M+F) x M/(M+F)] + [F/(M+F) x F/(M+F)]

= (M^2+F^2)/[(M+F)^2]

Statement 1)

M=41 - INSUF

Statement 2)

The probability of selecting one male and one female student is 15/41.

=[M/(M+F) x F/(M+F)] = 15/41

=MF/[(M+F)^2] = 15/41

So, MF = 15X

[(M+F)^2] = 41X

Lets go back to our question stem:

(M^2+F^2)/[(M+F)^2]

= (M^2+F^2) = [(M+F)^2] - 2MF = 41X - 2(15X) = 11X

So;

(M^2+F^2)/[(M+F)^2] = 11X/41X = 11/41

SUFF

Hence, B

Does my solution make sense ?
LGTCH
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by cramya » Sat Nov 15, 2008 2:54 pm
B is right but the I remember seeing the prob value to be 1-15/41 = 26/41

I will check but good approach.
The probability of selecting one male and one female student is 15/41.

=[M/(M+F) x F/(M+F)] = 15/41

=MF/[(M+F)^2] = 15/41
I am thinking this will be

[M/(M+F) x F/(M+F)] + [F/(M+F) x M/(M+F)] = 15/41
(Pick a male first then female or female first and then male)

It might get a bit complicated with this...Thoughts welcome.
Last edited by cramya on Sat Nov 15, 2008 3:12 pm, edited 3 times in total.
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by logitech » Sat Nov 15, 2008 3:05 pm
cramya wrote:B is right but the I remember seeing the prob value to be 1-15/41 = 26/41

I will check but good approach.
That's absolutely true! I am trying to find where I made a mistake.

Three ways: MM, FF, MF

p(MM) + p(FF) +p(MF) = 1

Perfect GMAT solution by the way.
LGTCH
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by cramya » Sat Nov 15, 2008 3:07 pm
That's absolutely true! I am trying to find where I made a mistake.

Three ways: MM, FF, FM

I think u missed FM i.e

Sample space

Four ways: MM, FF, MF,FM
Last edited by cramya on Sat Nov 15, 2008 3:25 pm, edited 1 time in total.
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by logitech » Sat Nov 15, 2008 3:20 pm
cramya wrote:
That's absolutely true! I am trying to find where I made a mistake.

Three ways: MM, FF, MF

I think u missed MM i.e

Sample space

Four ways: MM, FF, MF,MM
I don't follow you. We have two MM's and only one FF ?
LGTCH
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by cramya » Sat Nov 15, 2008 3:27 pm
I don't follow you. We have two MM's and only one FF ?
Oops typo with all these M's and F's. Corrected it.
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