Nutritional Information In a sample of 1000 U.S. adults,

Step 1 of 4

(a) In a sample of 1000 U.S. adults, 150 said they are very confident in the nutritional information on restaurant menus.

Four U.S. adults are selected at random without replacement.

We are asked to find the probability that all four adults are very confident in the nutritional information on restaurant menus.

Let E denote the event that U.S. adults are very confident in the nutritional information on restaurant menus.

Hence the probability that the U.S. adults are very confident in the nutritional information on restaurant menus is,

\(P(E)=\frac{150}{1000}=0.15\)

Since the individual probability is 0.15

Using the independence rule of probability, we can find the probability that all four adults are very confident in the nutritional information on restaurant menus.

\(P(\text { all four adults })=\frac{150}{1000} \times \frac{149}{999} \times \frac{148}{998} \times \frac{147}{997}=0.000489\)

Hence the probability is 0.000489