Whilst shopping, the probability that Caroline buys fruit is 0.7. The probability that Caroline buys a CD is 0.6.
Buying fruit and buying a CD are independent of each other.
Work out the probability that she only buys fruit, only buys a CD, or buys both.

Probability that Caroline buys fruit, a CD or both is 0.76.

Step-by-step explanation:

Let event A = Caroline buys fruit, event B = Caroline buys CD, Ac and Bc are complementary events.

Events AB, ABc, AcB and AcBc are jointly exhaustive and disjoint, hence P(AB) + P(ABc) + P(AcB) +P(AcBc) =1.

Events A and B independent, hence Ac and Bc independent too and probability P(AcBc) = P(Ac)*P(Bc) = (1 - P(A))(1-P(B)) = 0.6*0.4 = 0.24.

Required probability P(AB + ABc + AcB ) = P(AB) + P(ABc) + P(AcB) = 1- P(AcBc) = 1 - 0.24 = 0.76.

0.88

Step-by-step explanation:

P(FRUIT)=0.7

P(CD)0.6

0.7x0.6= 0.42

0.7x0.4=0.28

0.3x0.6=0.18

0.3x0.4=0.12

0.42+0.28+0.18= 0.88

d? maybe

step-by-step explanation:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!