Internet Doctors: Probability based on the survey data given in Exercise 1, assume that 55% of adults learn about medical symptoms more often from the internet than from their doctors.

a.Find the probability that three randomly selected adults all learn about medical symptoms more often from the internet than from their doctors.

b. Find the probability that three adults are randomly selected, at least one of them learns about medical symptoms more often from the internet than from their doctor.

It is known that 55% of adults learn about medical symptoms often from the internet than from their doctors.

A discrete random variable would follow binomial distribution when the number of successes is discussed in fixed trials with fixed probability of success in independent trials.

A discrete random variable X has probabilities as given below if it has binomial distribution for fixed trials n,

\(P\left( x \right) = \left\{ \begin{aligned}{l}{}^n{C_x}{p^x}{\left( {1 - p} \right)^{n - x}};x = 0,1,2,3,...,n\\0\,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;;Otherwise\end{aligned} \right.\)

a.

Let X be a random variable denoting the number of adults in 3 randomly selected adults who learn about medical symptoms more often from the internet than from their doctors.

As each person is independent and the probability for each person is same to gain information from the internet than the doctor, then X has binomial distribution with\(n = 3,p = 0.55\).

The proability of the event that all three adults learn about medical symptoms more often from the internet than from their doctors.

\(\begin{aligned}{c}P\left( {X = 3} \right) = {}^3{C_3}{\left( {0.55} \right)^3}{\left( {1 - 0.55} \right)^{3 - 3}}\\ = 1 \times {\left( {0.55} \right)^3}\\ = 0.166\end{aligned}\)

Thus, the probability that all three randomly selected adults learn about medical symptoms from the internet is 0.166.

b.

The probability that at least one of the three adults learn about medical symptoms more often from the internet than from their doctors is difference of probability of no success from 1.

It is computed as,

\(\begin{aligned}{c}P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\ = 1 - P\left( {X = 0} \right)\\ = 1 - {}^3{C_0}{\left( {0.55} \right)^0}{\left( {1 - 0.55} \right)^{3 - 0}}\\ = 1 - 0.091\\ = 0.909\end{aligned}\)

Thus, the probability thatat least one of the three randomly selected adults learn about medical symptoms from the internet is 0.909.