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Probability with discrete random distribution
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09 Oct 2018, 08:01
09 Oct 2018, 08:01
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Question Stats:
53% (01:00) correct
46% (02:14) wrong based on 28 sessions
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Hi Guys,
Does anybody know how to solve this question?
A class with 10 students will be split into five teams of two students each. There are four women and six men in the class. What is the probability that the first team chosen will have two men?
Thanks!!
Does anybody know how to solve this question?
A class with 10 students will be split into five teams of two students each. There are four women and six men in the class. What is the probability that the first team chosen will have two men?
Show: :: OA
0.8791
Thanks!!
Re: Probability with discrete random distribution
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09 Oct 2018, 09:27
09 Oct 2018, 09:27
Expert Reply
Hi,
please post the question under the right sub-forum. This is a numeric entry question, not a multiple ones.
Thank you so much
please post the question under the right sub-forum. This is a numeric entry question, not a multiple ones.
Thank you so much
Re: Probability with discrete random distribution
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09 Oct 2018, 12:10
09 Oct 2018, 12:10
4
1
Expert Reply
Could you share the source please
The question asks about choosing 2 ppl from 10 people and the probability of the two being male.
There are 4 women and 6 men.
The number of ways we can select 2 men = \(C_2^{6}=\frac{6!}{2! \times 4!}=15\).
The number of ways we can select 2 people = \(C_2^{10}=\frac{10!}{2! \times 8!}=45\).
So the probability of selecting 2 men = \(\frac{15}{45}=0.333\)
The question asks about choosing 2 ppl from 10 people and the probability of the two being male.
There are 4 women and 6 men.
The number of ways we can select 2 men = \(C_2^{6}=\frac{6!}{2! \times 4!}=15\).
The number of ways we can select 2 people = \(C_2^{10}=\frac{10!}{2! \times 8!}=45\).
So the probability of selecting 2 men = \(\frac{15}{45}=0.333\)
