Acheck of dorm rooms on a large college campus revealed that 34 % had refrigerators, 51 % had tvs, and 21 % had both a refrigerator and a tv. what's the probability that a randomly selected dorm room has

Consider the complete question is,

"A check of dorm rooms on a large college campus revealed that 34% had refrigerators, 51% had TVs, and 21% had both a TV and a refrigerator. what's the probability that a randomly selected dorm room has

a.a TV but no refrigerator?

b.a TV or a refrigerator, but not both?

c.neither a TV nor a refrigerator?"

Solution : Suppose A = event of having refrigerators,

B = event of having a TV,

We have,

P(A) = 34% = 0.34,

P(B) = 51% = 0.51,

P(A∩B) = 21% = 0.21

a. Probability of a TV but no refrigerator = P(A∩B')

= P(A) - P(A∩B)

= 0.34 - 0.21

= 0.31

b. Probability of a TV or a refrigerator, but not both = P(A∪B) - P(A∩B)

= P(A) + P(B) - P(A∩B) - P(A∩B)

= P(A) + P(B) - 2P(A∩B)

= 0.34 + 0.51 - 2(0.21)

= 0.43

c. Probability of neither a TV nor a refrigerator = P(A' ∩ B')

= 1 - P(A∪B)

= 1 - P(A) - P(B) + P(A∩B)

= 1 - 0.34 - 0.51 + 0.21

= 0.36

0,0

step-by-step explanation: