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# Three men are seeking for public office. Candidate A and B are given

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Three men are seeking for public office. Candidate A and B are given [#permalink]

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03 Jul 2010, 12:54
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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?
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Re: Three men are seeking for public office. Candidate A and B are given [#permalink]

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03 Jul 2010, 14:22
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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: Three men are seeking for public office. Candidate A and B are given [#permalink]

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03 Jul 2010, 14:41
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Expert's post
jovic1104 wrote:
Hello to all Experts in Statistics/Probabilities

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: Three men are seeking for public office. Candidate A and B are given [#permalink]

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03 Jul 2010, 19:23
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jovic1104 wrote:
Hello to all Experts in Statistics/Probabilities

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 --

5 girls: $$\frac{1}{3003}$$
5 boys: $$\frac{12}{143}$$
4 boys, 1 girl: $$\frac{50}{143}$$
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Re: Three men are seeking for public office. Candidate A and B are given [#permalink]

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03 Jul 2010, 20:51
Re: Three men are seeking for public office. Candidate A and B are given   [#permalink] 03 Jul 2010, 20:51
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