Probability Distribution Methods

Question

Q) Three towns measure the average amount of bread purchased per person per week. The probability that towns A, B and C measure the highest are 0.41, 0.36 and 0.23 respectively. Find the probability that over a five week period, town A records the highest in more weeks than the other towns.

I know that A would have to have the highest for at least 3 weeks (so either 3, 4 or all 5 weeks), so I did:

0.41^3 + 0.41^4 + 0.41^5 but this didn’t get me the right answer. What have I done wrong ?

Answer 1

Let's try to formalise this a bit.

$$P(A) = 0.41\\ P(B) = 0.36\\ P(C) = 0.23$$

Then, in order to get $A$ exactly 3 times, you would need to either

• Get $A$ 3 times, $B$ twice and $C$ $0$ times which has probability $\binom{5}{3}P(A)^3\cdot P(B)^2$
• Get $A$ 3 times, $B$ once and $C$ once which has probability $\binom{5}{3}P(A)^3\cdot \binom{2}{1}P(B)\cdot P(C)$
• Get $A$ 3 times, $C$ twice and $B$ $0$ times which has probability $\binom{5}{3}P(A)^3\cdot P(C)^2$

If you add these up, you get the probability for $A$ to appear exactly 3 times. Which would be $\binom{5}{3}P(A)^3\cdot (1-P(A))^2$ which is the binomial distribution.

Use a similar approach to find the probability that $A$ appears exactly 4 times and exactly 5 times.

The end result should be $\binom{5}{3}P(A)^3\cdot (1-P(A))^2 + \binom{5}{4}P(A)^4\cdot (1-P(A)) + P(A)^5$.