A recent survey indicated that 82% of single women aged 25 years old will be married in their
lifetime. Using the binomial distribution find the probability that two or three women in a
sample of twenty will never be married. Let X be the number of women in the sample never married.
We can use the binomial distribution to calculate the probability since there are two possible outcomes (married or not married) and the probability of success (not married) remains constant for each trial. The probability of failure (married) is $p = 1 - 0.82 = 0.18$. The number of trials is $n = 20$, and the number of successes we want to find the probability for is $x = 2$ or $x = 3$.

The probability mass function for the binomial distribution is:

$$P(X=x) = {n \choose x} p^x (1-p)^{n-x}$$

First, let's find $P(X=2)$:

$${20 \choose 2} (0.18)^2 (1-0.18)^{20-2} \approx 0.275$$

Next, let's find $P(X=3)$:

$${20 \choose 3} (0.18)^3 (1-0.18)^{20-3} \approx 0.223$$

Finally, to find the probability that two or three women in the sample will never be married, we can add these probabilities:

$$P(X=
  To use the binomial distribution, we need to assume that the probability of a woman getting married is independent of the other women in the sample. Let's use p as the probability of a woman getting married.

The number of women in the sample who will never be married can be modeled as a binomial random variable X with n = 20 and p = 0.82 (the probability that a woman aged 25 will be married).

The probability of exactly k women in the sample never being married can be found using the binomial distribution formula:

P(X = k) = (n! / k! (n - k)!)(p^k \* (1 - p)^(n - k))

where k = 0, 1, 2,..., n.

To find the probability that two or three women in the sample will never be married, we can use the following inequalities:

P(X = 2) = (20! / 2! (20 - 2)!)(0.82^2 \* (1 - 0.82)^(20 - 2)) ≈ 0.28

P(X = 3) = (20! / 3! (20 - 3)!)(0.82^3 \* (1 - 0.