Question

The results of a survey in the cafeteria show that 20% of students like ketchup, 50% like mustard, and 15% like both ketchup and mustard. What is the probability that a student likes neither ketchup nor mustard? A) 25% B) 45% C) 75% D) 85%

Answer 1

B) 45%

The probability that neither K nor M occurs = P(K⁻∩M⁻) = 0.45 = 45%

Explanation:

Explanation:-

Given data the results of a survey in the cafeteria show that 20% of students like ketchup.

Let "K" be the event of students like ketchup

P(K) = 20% = 0.20

Given data the results of a survey in the cafeteria show that 50% of students like mustard.

Let 'M' be the event of students like mustard

P(M) = 50% =0.50

Let 15% like both ketchup and mustard

P(K∩M) = 0.15

The probability that neither K nor M occurs = P(K⁻∩M⁻)

= P(S - (K∪M)

= P(S) - P(K∪M)

Total sample space P(S) =1

= 1 - (P(K) +P(M) -P(K∩M)

= 1 - (0.20 + 0.50 - 0.15)

= 1 - 0.55 = 0.45

Final answer:-

The probability that neither K nor M occurs = P(K⁻∩M⁻) = 0.45