Question

Suppose that 65% of all adults regularly consume coffee, 60% regularly consume carbonated soda, and 75% regularly consume at least one of these two products. (a) What is the probability that a randomly selected adult regularly consumes both coffee and soda? (b) What is the probability that a randomly selected adult doesn't regularly consume at least one of these two products?

Answer 1

(a) 0.50 or 50%

(b) 0.25 or 25%

Explanation:

We are given that;

Probability of adults that regularly consume coffee, P(C) = 0.65

Probability of adults that regularly consume carbonated soda, P(S) = 0.60

Probability of adults that regularly consume at least one of these two products, P(C
`\bigcup` S) = 0.75

(a) Probability that a randomly selected adult regularly consumes both coffee and soda is given by P(C
`\bigcap` S) ;

P(C
`\bigcup` S) = P(C) + P(S) - P(C
`\bigcap` S)

0.75 = 0.65 + 0.60 - P(C
`\bigcap` S)

P(C
`\bigcap` S) = 1.25 - 0.75 = 0.50

Therefore, the probability that a randomly selected adult regularly consumes both coffee and soda is 0.50 or 50%.

(b) Probability that a randomly selected adult doesn't regularly consume at least one of these two products = P(C
`\bigcup` S)'

P(C
`\bigcup` S)' = 1 - P(C
`\bigcup` S) = 1 - 0.75 = 0.25 or 25% .

Answer 2

Answer: a) 50%, b) 25%.

Explanation:

Since we have given that

Probability of all adults regularly consume coffee P(C) = 65%

Probability of all adults regularly consume soda P(S) = 60%

Probability of all adults regularly either one of them P(C ∪S) = 75%

According to question, we get that

(a) What is the probability that a randomly selected adult regularly consumes both coffee and soda?

`P(C\cup S)=P(C)+P(S)-P(C\cap S)\\\\0.75=0.65+0.60-P(C\cap S)\\\\0.75=1.25-P(C\cap S)\\\\P(C\cap S)=1.25-0.75=0.5=50\%`

(b) What is the probability that a randomly selected adult doesn't regularly consume at least one of these two products?

P(C∪S)'=1-P(C∪S)

`P(C\cup S)'=1-0.75=0.25=25\%`

Hence, a) 50%, b) 25%.