Question

n 2019, approximately 97.4% of all the runners who started the Boston Marathon (in Boston, Massachusetts, USA) were able to complete the 42.2 km (26.2 mile) race. If 100 runners are chosen at random, find the probability that at least 5 of them did not finish the marathon

Answer 1

Answer:

0.1199 = 11.99% probability that at least 5 of them did not finish the marathon

Step-by-step explanation:

For each runner, there are only two possible outcomes. Either they finished the marathon, or they did not. The probability of a runner completing the marathon is independent of any other runner. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

97.4% finished:

This means that 100 - 97.4 = 2.6% = 0.026 did not finish, which means that $p = 0.026$

100 runners are chosen at random

This means that $n = 100$

Find the probability that at least 5 of them did not finish the marathon

This is:

$P(X \geq 5) = 1 - P(X < 5)$

In which

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{100,0}.(0.026)^{0}.(0.974)^{100} = 0.0718$

$P(X = 1) = C_{100,1}.(0.026)^{1}.(0.974)^{99} = 0.1916$

$P(X = 2) = C_{100,2}.(0.026)^{2}.(0.974)^{98} = 0.2531$

$P(X = 3) = C_{100,3}.(0.026)^{3}.(0.974)^{97} = 0.2207$

$P(X = 4) = C_{100,4}.(0.026)^{4}.(0.974)^{96} = 0.1429$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0718 + 0.1916 + 0.2531 + 0.2207 + 0.1429 = 0.8801$

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.8801 = 0.1199$

0.1199 = 11.99% probability that at least 5 of them did not finish the marathon