Question

A study determined that 6% of children under 18 years old lived with their father only. Find the probability that none of the children selected at random from 11 children under 18 yearsold lived with their father onlyThe probability that none of the children ived with their father only is(Do not found until the final answer. Then found to the nearest thousandth as needed)

Answer 1

The question given is under binomial distribuion. The binomial probability distribution is a discrete probability distribution or based on a discrete Bernoulli random variable,

X∼B(n, p)

Where n is the number of trials and p is the probability of success. The binomial distribution is used when the trials are independent and finite with the same probability of success.

The binomial distribution law is given by the formula below:

`P(X=x)=^nC_rp^r(1-p)^(n-r)`

From the question given

`\begin{gathered} p=6\text{ \%=}(6)/(100)=0.06 \\ 1-p=1-0.06=0.94 \\ n=11 \end{gathered}`

The probability that none of the children ived with their father only is

`P(X=0)`
`\begin{gathered} P(X=0)=^(11)C_0p^0(1-p)^(11-0) \\ P(X=0)=^(11)C_o*0.06^0*0.94^(11) \\ ^(11)C_0=(11!)/(0!*(11-0)!)=(11!)/(1*11!)=1 \end{gathered}`
`\begin{gathered} P(X=0)=1*1*0.506298 \\ P(X=0)=0.506298 \\ P(X=0)\approx0.506(nearest\text{ thousandth)} \end{gathered}`

Hence, probability that none of the children ived with their father only is 0.506 to the nearest thousandth