Question

It is claimed that 55% of marriages in the state of California end in divorce within the first 15 years. A large study was started 15 years ago and has been tracking hundreds of marriages in the state of California. Suppose 10 marriages are randomly selected. What is the probability that less than two of them ended in a divorce

Answer 1

Answer:

0.0045 = 0.45% probability that less than two of them ended in a divorce

Step-by-step explanation:

For each marriage, there are only two possible outcomes. Either it ended in divorce, or it did not. The probability of a marriage ending in divorce is independent of any other marriage. 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.

55% of marriages in the state of California end in divorce within the first 15 years.

This means that $p = 0.55$

Suppose 10 marriages are randomly selected.

This means that $n = 10$

What is the probability that less than two of them ended in a divorce?

This is

$P(X < 2) = P(X = 0) + P(X = 1)$

In which

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

$P(X = 0) = C_{10,0}.(0.55)^{0}.(0.45)^{10} = 0.0003$

$P(X = 1) = C_{10,1}.(0.55)^{1}.(0.45)^{9} = 0.0042$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0003 + 0.0042 = 0.0045$

0.0045 = 0.45% probability that less than two of them ended in a divorce