Question

a) An athlete is running in four races and in each race she has a 60% chance of
winning. What is the probability that she will win at least two races?
b) A website has on the average two hits per hour. Assuming a Poisson distribution for
the number of hits per hour (X), calculate the probability that there are at most three
hits.

Answer 1

Answer on Question #69782 – Math – Statistics and Probability

Question

a) An athlete is running in four races and in each race she has a 60% chance of winning. What is the probability that she will win at least two races?

Solution

We have binomial distribution with

$$p = 0.6 \text{ and } n = 4.$$ $$P(X \geq 2) = 1 - P(X = 0) - P(X = 1)$$ $$P(X \geq 2) = 1 - \frac{4!}{0!(4 - 0)!} (0.6)^0 (1 - 0.6)^{4 - 0} - \frac{4!}{1!(4 - 1)!} (0.6)^1 (1 - 0.6)^{4 - 1}$$ $$P(X \geq 2) = 1 - (0.4)^4 - 4(0.6)(0.4)^3 = 0.8208$$

Question

b) A website has on the average two hits per hour. Assuming a Poisson distribution for the number of hits per hour (X), calculate the probability that there are at most three hits.

Solution

$$\lambda = 2.$$ $$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$$ $$P(X \leq 3) = e^{-2} \frac{2^0}{0!} + e^{-2} \frac{2^1}{1!} + e^{-2} \frac{2^2}{2!} + e^{-2} \frac{2^3}{3!}$$ $$P(X \leq 3) = 0.8571.$$

Answer: a) 0.8208; b) 0.8571.

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