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Out of 5 boys and 6 girls, 4 students are sellected at rando

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Out of 5 boys and 6 girls, 4 students are sellected at rando  [#permalink]

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New post 25 Aug 2018, 02:03
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Question Stats:

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Out of 5 boys and 6 girls, 4 students are selected at random, what is the probability that selection has more number of girls than the number of boys?

\(A) \frac{1}{33}\)
\(B) \frac{15}{33}\)
\(C) \frac{23}{66}\)
\(D) \frac{12}{33}\)
\(E) \frac{1}{3}\)

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Out of 5 boys and 6 girls, 4 students are sellected at rando  [#permalink]

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New post 25 Aug 2018, 02:24
hasnain3047 wrote:
Out of 5 boys and 6 girls, 4 students are selected at random, what is the probability that selection has more number of girls than the number of boys?

\(A) \frac{1}{33}\)
\(B) \frac{15}{33}\)
\(C) \frac{23}{66}\)
\(D) \frac{12}{33}\)
\(E) \frac{1}{3}\)

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OA: C

Total Number of case of selecting four students out of 11 \(= C(11,4) = \frac{11!}{7!4!}=330\)

Number of ways of selecting 4 girls or 3 girls and 1 boy to form a 4 member team \(= C(6,4)*1 + C(6,3)*C(5,1)=\frac{6!}{4!2!}*1 +\frac{6!}{3!3!}*\frac{5!}{4!1!}=115\)

Required probability \(= \frac{115}{330} =\frac{23}{66}\)
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Re: Out of 5 boys and 6 girls, 4 students are sellected at rando  [#permalink]

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New post 25 Aug 2018, 02:30

Solution



Given:
    • Out of 5 boys and 6 girls, 4 students are selected at random.

To find:
    • The probability that selection has more number of girls than the number of boys.

Approach and Working:
From 11 students, 4 can be selected in \(^{11}C_4\) = 330 ways

Now, if number of girls are more than boys, then the possible cases are:
    • 3 girls and 1 boy = \(^6C_3\) x \(^5C_1\) = 20 x 5 = 100
    • 4 girls and 0 boy = \(^6C_4\) = 15
    • Hence, the probability = \(\frac{100+15}{330} = \frac{115}{330} = \frac{23}{66}\)

Hence, the correct answer is option C.

Answer: C

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Out of 5 boys and 6 girls, 4 students are sellected at rando  [#permalink]

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New post 28 Aug 2018, 01:32
hasnain3047 wrote:
Out of 5 boys and 6 girls, 4 students are selected at random, what is the probability that selection has more number of girls than the number of boys?

\(A) \frac{1}{33}\)
\(B) \frac{15}{33}\)
\(C) \frac{23}{66}\)
\(D) \frac{12}{33}\)
\(E) \frac{1}{3}\)

Posted from my mobile device



more number of girls than boys is possible in 2 cases

1. 3 girls and 1 boy = \(\frac{6c3*5}{11c4}\)
2. all 4 girls = \(\frac{6c4}{11c4}\)

= \(\frac{115}{330}\) =\(\frac{23}{66}\)
Out of 5 boys and 6 girls, 4 students are sellected at rando &nbs [#permalink] 28 Aug 2018, 01:32
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