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Mr. Yutaro’s class contains 5 boys and 11 girls. If two students are

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prasadrg
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Mr. Yutaro’s class contains 5 boys and 11 girls. If two students are [#permalink] New post   23 Jan 2010, 15:49
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Mr. Yutaro’s class contains 5 boys 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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C
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BrentGMATPrepNow
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Re: Mr. Yutaro’s class contains 5 boys and 11 girls. If two students are [#permalink] New post   28 Dec 2017, 16:32
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prasadrg wrote:
Mr. Yutaro’s class contains 5 boys 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
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andrewcs
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Re: Mr. Yutaro’s class contains 5 boys and 11 girls. If two students are [#permalink] New post   23 Jan 2010, 16:03
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prasadrg 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


Probability of a boy being chosen first, then a girl:
5/16 * 11/15

Probability of a girl being chosen first, then a boy:
11/16 * 5/15

Probability of a boy and a girl being chosen:
(5*11+5*11) / (15*16)
=11+11/3*16
=22/48
=11/24, so C.
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Re: Mr. Yutaro’s class contains 5 boys and 11 girls. If two students are [#permalink] New post   19 Feb 2010, 05:40
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(5C1*11C1)/16C2 = 11/24
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Re: Mr. Yutaro’s class contains 5 boys and 11 girls. If two students are [#permalink] New post   19 Feb 2010, 12:24
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umeshkathuria wrote:
(5C1*11C1)/16C2 = 11/24


Actually, problems of this kind, prefer subtracting from all possible probabilities = 1.

Subtract probability of choosing only two boys and probability of choosing only two girls and that should give the final answer. This will be more helpful when you have many more selections that happen and not just 2 and one from each is required.
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Re: Mr. Yutaro’s class contains 5 boys and 11 girls. If two students are [#permalink] New post   20 Feb 2010, 08:39
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prasadrg 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


Ways to choose one boy = 5c1 = 5
Ways to choose one girl = 11c1 = 11

Total ways = 16c2 = 16*15

p=(5*11)/16*15 - 11/24 hence C
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Re: Mr. Yutaro’s class contains 5 boys and 11 girls. If two students are [#permalink] New post   28 Dec 2017, 19:10
Number of ways of selecting two students from the class of 5 boys & 11 girls = 16C2 (total) =15*16/2
Number of ways of selecting 1 boy & 1 girl = 5C1 * 11C1 = 5*11
probability = 5*11/15*8 = 11/24
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Re: Mr. Yutaro’s class contains 5 boys and 11 girls. If two students are [#permalink] New post   21 Sep 2019, 12:20
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prasadrg wrote:
Mr. Yutaro’s class contains 5 boys 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


The probability that a girl and a boy, in that order, are chosen is 11/16 x 5/15 = 11/16 x 1/3 = 11/48. The probability that a boy and a girl, in that order, are chosen is also 11/48. Therefore, the probability that a boy and a girl are chosen is 11/48 + 11/48 = 22/48 = 11/24.

Answer: C
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Re: Mr. Yutaro’s class contains 5 boys and 11 girls. If two students are [#permalink] New post   07 Dec 2020, 13:20
2
1 x 5/16 x 11/15 = 110/240 = 11/24

2
1 indicates that there's two possibilities to choose one girl and one boy, e.g. 5/16 x 11/15 or 11/16 x 5/15

2
1 is combinatorial, and in this case it will just be (2!)/(1!x1!) = 2
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Mr. Yutaro’s class contains 5 boys and 11 girls. If two students are [#permalink] New post   18 Feb 2021, 07:50
(5/16)(11/15)*2! = 11/24
boy girl

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Re: Mr. Yutaro’s class contains 5 boys and 11 girls. If two students are [#permalink] New post   18 Feb 2021, 08:07
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prasadrg wrote:
Mr. Yutaro’s class contains 5 boys 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


1. First boy and than Girl

=\(\frac{5}{16} * \frac{11}{15}\)

=\(\frac{11}{48}\)

2. First Girl and than Boy

=\(\frac{11}{48}\)

Total = \(\frac{11}{48} + \frac{11}{48}\)

= \(\frac{11}{24}\)

Ans C
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Re: Mr. Yutaros class contains 5 boys and 11 girls. If two students are [#permalink] New post   06 Dec 2022, 12:23
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