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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?
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?
whiplash2411
Joined: 09 Jun 2010
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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 = \(\frac{Number of ways of 4 boys and 1 girl}{Total ways of picking}\) = \(\frac{1050}{3003}\)
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 = \(\frac{Number of ways of 4 boys and 1 girl}{Total ways of picking}\) = \(\frac{1050}{3003}\)
03 Jul 2010, 14:41
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jovic1104
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.
