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Probability questions need help.

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Probability questions need help. [#permalink] New post 03 Jul 2010, 11:54
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Hello to all Experts in Statistics/Probabilities

Please help me to solve the 2 probabilities questions below.

Q1. Three men are seeking for public office. Candidate A and B are given about the same chance of winning but Candidate C is given the chance twice as either A or B. What is the probability that A does not win? What is the probability of B?

Q2. In a certain children school, a teacher randomly selects 5 pre-schoolers from a class consisting of 10 boys and 5 girls. What is the probability of getting all 5 girls? 5 boys? Four boys and 1 girl?

It will be great if detailed explanation to the solution is provided. Many thanks in advance.
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Re: Probability questions need help. [#permalink] New post 03 Jul 2010, 13:22
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For the first question:

Let us assume the following:

P(A) = x
P(B) = x
P(C) = 2x

Sum of probabilities = 1

4x=1

x= \frac{1}{4}

From there P(A) = P(B) =x= \frac{1}{4}

P(A not winning) = 1-P(A) = 1- \frac{1}{4} = \frac{3}{4}

For the second question, I am not sure if the solution is right, but here's my take on it:

Total ways of picking 5 students = 15C5 = \frac{15!}{10!*5!} = 3003

Number of ways of picking 5 girls = 5C5 = 1

So, probability of 5 girls = \frac{Number of ways of girls}{Total ways of picking} = \frac{1}{3003}

Number of ways of picking 5 boys = 10C5 = \frac{10!}{5!*5!} = 252

Probability of 5 boys = \frac{Number of ways of girls}{Total ways of picking} = \frac{252}{3003}

Number of ways of picking 4 boys and 1 girl = 10C4*5C1 = \frac{10!}{4!*6!}*\frac{5!}{4!*1!}= 1050

Probability = \frac{Number of ways of 4 boys and 1 girl}{Total ways of picking} = \frac{1050}{3003}
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Re: Probability questions need help. [#permalink] New post 03 Jul 2010, 13:41
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jovic1104 wrote:
Hello to all Experts in Statistics/Probabilities

Please help me to solve the 2 probabilities questions below.

Q1. Three men are seeking for public office. Candidate A and B are given about the same chance of winning but Candidate C is given the chance twice as either A or B. What is the probability that A does not win? What is the probability of B?

Q2. In a certain children school, a teacher randomly selects 5 pre-schoolers from a class consisting of 10 boys and 5 girls. What is the probability of getting all 5 girls? 5 boys? Four boys and 1 girl?

It will be great if detailed explanation to the solution is provided. Many thanks in advance.


Q1. Three men are seeking for public office. Candidate A and B are given about the same chance of winning but Candidate C is given the chance twice as either A or B. What is the probability that A does not win? What is the probability of B?

As there are only 3 candidates and assuming that either one of them must win, then their chances of winning must add up to 100%. So if chances of winning of A is x%, then chances of winning of B will also be x% and of C 2x%. Hence x+x+2x=100% --> x=25%=\frac{1}{4}.

So chances of B winning is x=\frac{1}{4}

Now, if chances of A winning is \frac{1}{4}, then chances of not winning will be 1-\frac{1}{4}=\frac{3}{4}.

Q2. In a certain children school, a teacher randomly selects 5 pre-schoolers from a class consisting of 10 boys and 5 girls. What is the probability of getting all 5 girls? 5 boys? Four boys and 1 girl?

Probability=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes}

A. \frac{C^5_5}{C^5_{15}}, where C^5_5 is the # of ways to choose 5 girls out of 5 and C^5_{15} is the total # of ways to choose any 5 children out of total 15.

B. \frac{C^5_{10}}{C^5_{15}}, where C^5_{10} is the # of ways to choose 5 boys out of 5 and C^5_{15} is the total # of ways to choose any 5 children out of 15.

C. \frac{C^4_{10}*C^1_{5}}{C^5_{15}}, where C^4_{10} is the # of ways to choose 4 boys out of 5, C^1_5 is the # of ways to choose 1 girls out of 5 and C^5_{15} is the total # of ways to choose any 5 children out of 15.

Hope it helps.
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Re: Probability questions need help. [#permalink] New post 03 Jul 2010, 18:23
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jovic1104 wrote:
Hello to all Experts in Statistics/Probabilities

Please help me to solve the 2 probabilities questions below.

Q1. Three men are seeking for public office. Candidate A and B are given about the same chance of winning but Candidate C is given the chance twice as either A or B. What is the probability that A does not win? What is the probability of B?

Q2. In a certain children school, a teacher randomly selects 5 pre-schoolers from a class consisting of 10 boys and 5 girls. What is the probability of getting all 5 girls? 5 boys? Four boys and 1 girl?

It will be great if detailed explanation to the solution is provided. Many thanks in advance.



Just to save some people some time --

Q2: the reduced answers are:

5 girls: \frac{1}{3003}
5 boys: \frac{12}{143}
4 boys, 1 girl: \frac{50}{143}
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Re: Probability questions need help. [#permalink] New post 03 Jul 2010, 19:51
Thanks to all your reply. Your answer and explanation really makes sense. Kudos points to all of you.
Re: Probability questions need help.   [#permalink] 03 Jul 2010, 19:51
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