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Suppose that 55% of all adults regularly consume coffee,

45% regularly consume carbonated soda, and 70% regularly

consume at least one of these two products.

AWhat is the probability that a randomly selected adult

regularly consumes both coffee and soda?

BWhat is the probability that a randomly selected adult

doesn’t regularly consume at least one of these two

products?

Verified Answer :

Answer: The probability that a randomly selected adult doesn’t regularly consume at least one of these two products is 30%

Be:
A: Adults regularly consume coffee
B: Adults regularly consume carbonated soda
C: Adults doesn't regularly consume at leas one of these two products.

55% of all adults regularly consume coffee→P(A)=55%=55/100→P(A)=0.55
45% regularly consume carbonated soda→P(B)=45%=45/100→P(B)=0.45
70% regularly consume at least one of these two products→P(A U B)=70%=70/100→P(A U B)=0.70

a) What is the probability that a randomly selected adult regularly consumes both coffee and soda?
P(A ∩ B)=?

P(A U B)=P(A) + P(B) - P(A ∩ B)

Replacing P(A U B)=0.70; P(A)=0.55; and P(B)=0.45 in the equation above:
0.70=0.55 + 0.45 - P(A ∩ B)→
0.70=1.00 - P(A ∩ B)

Solving for P(A ∩ B): Subtracting 1.00 both sides of the equation:
0.70-1.00=1.00 - P(A ∩ B) -1.00→
-0.30 = - P(A ∩ B)

Multiplying both sides of the equation by (-1):
(-1)(-0.30) = (-1)[ - P(A ∩ B)]→
0.30 = P(A ∩ B)→
P(A ∩ B) =0.30→
P(A ∩ B) = (0.30)100%→
P(A ∩ B) = 30%

Answer: The probability that a randomly selected adult regularly consumes both coffee and soda is 70%

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

P(A U B) + P(C)=1

Replacing P(A U B)= 0.70 in the equation above:
0.70+P(C)=1

Solving for P(C). Subtracting 0.70 both sides of the equation:
0.70+P(C)-0.70=1-0.70→
P(C)=0.30→
P(C)=(0.30)100%→
P(C)=30%

Answer: The probability that a randomly selected adult doesn’t regularly consume at least one of these two products is 30%