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

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.