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by thephoenix » Mon Feb 22, 2010 10:51 pm
Mr. Yutaro’s class contains 5boys and 11 girls. If two students are chosen, one at a time, from the class, what is
the probability that a boy and a girl are chosen?
A. 9/48
B. 9/20
C. 11/24
D. 27/48
E. 13/16
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by ajith » Mon Feb 22, 2010 11:02 pm
thephoenix wrote:Mr. Yutaro’s class contains 5boys and 11 girls. If two students are chosen, one at a time, from the class, what is
the probability that a boy and a girl are chosen?
A. 9/48
B. 9/20
C. 11/24
D. 27/48
E. 13/16
No of ways in which 2 students can be chosen from the group of 16 (11+5) = 16C2 = 16*15/2 = 8*15 =120
No of ways in which 1 girl and 1 boy can be chosen from 11 girls and 5 boys = 11C1*5C1 = 55
[spoiler]
Probability = 55/120 = 11/24[/spoiler]
Always borrow money from a pessimist, he doesn't expect to be paid back.
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by analyst218 » Fri Feb 26, 2010 1:33 pm
Can neone help me understand this more clearly.
I understand why ajith is correct.
However, can't you also say

P(A) = boys
P(B) = girls

P(AandB) = choosing boys and girls?


P(A)*P(BgivenA) = [5/16]*[11/15] = 11/48

I get 11/24 as above when I do P(B)*P(B) = 11/16 * 10/15 = 11/24

Correct me what I'm thinking wrong.....

thanks
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by arzanr » Fri Feb 26, 2010 8:39 pm
analyst218 wrote:Can neone help me understand this more clearly.
I understand why ajith is correct.
However, can't you also say

P(A) = boys
P(B) = girls

P(AandB) = choosing boys and girls?


P(A)*P(BgivenA) = [5/16]*[11/15] = 11/48

I get 11/24 as above when I do P(B)*P(B) = 11/16 * 10/15 = 11/24

Correct me what I'm thinking wrong.....

thanks
Analyst,

You're thinking is correct but you stopped halfway. The question says that a boy and girl is picked which means that you can either first pick a boy and then a girl or first pick a girl and then a boy. You calculated the probability of picking a boy first and then a girl which is equal to 11/24. Now add the prob of picking the girl first and you get:

11/48 + 11/48 = 22/48 = 11/24
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by analyst218 » Mon Mar 01, 2010 11:17 am
picking boy and girl= p(boy then girl) + p(girl then boy)
= P(boy)P(girlgivenboy) + p(girl)p(boygivengirl)
= 11/48 + 11/48 = 11/24

makes total sense...
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by Brent@GMATPrepNow » Thu Jan 09, 2020 6:16 am
thephoenix wrote:Mr. Yutaro’s class contains 5boys and 11 girls. If two students are chosen, one at a time, from the class, what is
the probability that a boy and a girl are chosen?
A. 9/48
B. 9/20
C. 11/24
D. 27/48
E. 13/16
P(1 boy and 1 girl chosen) = P(1st selection is boy AND 2nd selection is girl OR 1st selection is girl AND 2nd selection is boy)
= P(1st selection is boy AND 2nd selection is girl) + P(1st selection is girl AND 2nd selection is boy)
= P(1st selection is boy) x P(2nd selection is girl) + P(1st selection is girl) x P(2nd selection is boy)
= 5/16 x 11/15 + 11/16 x 5/15
= 55/240 + 55/240
= 110/240
= 11/24
= C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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