Question

A researcher wishes to conduct a study of the color preferences of new car buyers. Suppose that 50% of this population prefers the color blue. If 10 buyers are randomly selected what is the probability that exactly 3 buyers would prefer blue? Round to answer to four decimal places

Answer 1

0.1172

Explanation:

The probability that a buyer prefers blue = 50% = 0.5

The number of buyers =10

We want to find the probability that exactly 3 buyers would prefer blue.

To solve this problem, we use the binomial distribution formula given below:

`\begin{gathered} P(x=k)=\left(\begin{array}{l}n \\ k\end{array}\right)p^k(1-p)^(n-k) \\ =(n!)/(k!(n-k)!)p^k(1-p)^(n-k) \end{gathered}`

In the given case:

• n=10

,

• k=3

,

• p=0.5

Therefore:

`\begin{gathered} P(x=3)=(10!)/((10-3)!3!)(0.5)^3(1-0.5)^(10-3) \\ =(10!)/(7!3!)*(0.5)^3(0.5)^7 \\ =(10*9*8*7!)/(7!*6)*(0.5)^3*(0.5)^7 \\ =(10*9*8)/(6)*(0.5)^3*(0.5)^7 \\ \approx0.1172 \end{gathered}`

The probability that exactly 3 buyers would prefer blue is 0.1172 (rounded to four decimal places).