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%
Toda fraccion cuyo denominador (el numero de abajo) es multiplo de 10 es considerada una fraccion decimal. No importa si el numerador es mayor al denominador. Justamente el sistema decimal fue concebido para facilitar la multiplicacion y division de los numeros por algun multiplo de 10.