It has been estimated that 40% of marriages end in divorce. If you randomly select 9 recently married couples what is the probability that at least one of the marriages will end in divorce?

Step-by-step explanation:

Hello!

In this example you are interested to know how many matrimonies and up in divorce.

The study variable is X: number of marriages that ended up in divorce in a sample of nine.

This variable is a discrete variable and a quick check will tell you if it follows the binomial criteria, if it does, then the variable will have binomial distribution:

Binomial criteria:

1. The number of observation of the trial is fixed (n = 9)

2. Each observation in the trial is independent, this means that none of the trials will affect the probability of the next trial.

3. The probability of success in the same from one trial to another (the "success" of the trial is that the matrimony ended up in divorce and it's estimated probability is p=0.40

So X≈ Bi (n;ρ)

You have to calculate the probability that at least one of the marriages end in divorce, this is that "one or more" of the marriages end in divorce, symbolically:

P(X≥1)

This expression includes the probabilities P(X=1), P(X=2), P(X=3), P(X=4), P(X=5), P(X=6), P(X=7), P(X=8), P(X=9).

One way to calculate this is by applying the formula for calculating point probabilities under the binomial distribution nine times and then add them or you can use a table of the binomial distribution. These tables show cumulative probabilities for many distributions, P(X≤x₀)

To reach the probability value using the table you have to do the following:

P(X≥1)= 1 - P(X<1)

One is the max cumulative probability, so you have to subtract from it everything that isn't included in the expression, the only probability that is "left out" from P(X≥1) is P(X=0), in the table you have to look for the probability of P(X≤0)= 0.0101, then you calculate the probability as follow:

P(X≥1)= 1 - P(X<1)= 1 - P(X≤0)= 1 - 0.0101= 0.9899

I hope this helps!