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Out of 5 boys and 6 girls, 4 students are selected at random such that
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Updated on: 23 Feb 2024, 02:42
Updated on: 23 Feb 2024, 02:42
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Question Stats:
16% (02:18) correct
83% (02:24) wrong based on 12 sessions
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Out of 5 boys and 6 girls, 4 students are selected at random such that at least 1 boy and 1 girl is selected, what is the probability that the selection has more number of girls than the number of boys?
A. 1/33
B. 5/33
C. 10/33
D. 10/31
E. 21/31
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A. 1/33
B. 5/33
C. 10/33
D. 10/31
E. 21/31
Post A Detailed Correct Solution For The Above Questions And Get Kudos.
Question From Our Project: Free GRE Prep Club Tests in Exchange for 20 Kudos
\(\Longrightarrow\) GRE - Quant Daily Topic-wise Challenge
\(\Longrightarrow\) GRE Math Essentials - A most comprehensive handout!!
\(\Longrightarrow\) ALL the GRE Quant and Verbal Practice Directories in ONE place (2023)
\(\Longrightarrow\) Best GRE Test Books of 2023
Re: Out of 5 boys and 6 girls, 4 students are selected at random such that
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23 Feb 2024, 00:52
23 Feb 2024, 00:52
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it should be 10/31.
Out of 5 boys and 6 girls, 4 students are selected at random such that
[#permalink]
23 Feb 2024, 02:42
23 Feb 2024, 02:42
Expert Reply
Yes sir. It is correct. The table of the answers reported C. However is not possible. D must be the answer.
\(\frac{C^6_3 \times C^5_1}{C^{11}_4 - C^5_4 - C^6_4}= \frac{20 \times 5}{330-5-15}=\frac{100}{310}=\frac{10}{31}\)
Thank you
\(\frac{C^6_3 \times C^5_1}{C^{11}_4 - C^5_4 - C^6_4}= \frac{20 \times 5}{330-5-15}=\frac{100}{310}=\frac{10}{31}\)
Thank you
