Question

It is known that 10% of the 9-year old children in a town have three siblings. Fifteen 9-year old children are selected at random. Find the probability that more than 3 of these selected children have three siblings (round off to second decimal place).

Answer 1

Using the binomial distribution, it is found that there is a 0.0555 = 5.55% probability that more than 3 of these selected children have three siblings.

What is the binomial distribution formula?

The formula is:

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

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

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem, for the parameters, we consider that:

• 10% of the 9-year old children in a town have three siblings, hence p = 0.1.

• Fifteen 9-year old children are selected at random, hence n = 15.

The probability that more than 3 of these selected children have three siblings is given by:

$P(X > 3) = 1 - P(X \leq 3)$

In which:

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

Then:

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

$P(X = 0) = C_{15,0}.(0.1)^{0}.(0.9)^{15} = 0.2059$

$P(X = 1) = C_{15,1}.(0.1)^{1}.(0.9)^{14} = 0.3432$

$P(X = 2) = C_{15,2}.(0.1)^{2}.(0.9)^{13} = 0.2669$

$P(X = 3) = C_{15,3}.(0.1)^{3}.(0.9)^{12} = 0.1285$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.2059 + 0.3432 + 0.2669 + 0.1285 = 0.9445$

P(X > 3) = 1 - 0.9445 = 0.0555

0.0555 = 5.55% probability that more than 3 of these selected children have three siblings.

More can be learned about the binomial distribution at brainly.com/question/24863377