Question

A check of dorm rooms on a large college campus revealed that 48% had refrigerators, 43% had TVs, and 63% had at least a TV or a refrigerator.

What is the probability that a randomly selected dorm room has both a TV and a refrigerator?

a: 0.00 b:0.28 c:0.41 d:0.59 e:0.72

What is the probability that a randomly selected dorm room has either a TV or a refrigerator, but not both?

a: 0.00 b: 0.18 c: 0.35 d: 0.85 e: 0.99

What is the probability that a randomly selected dorm room has only TV but no refrigerator?

a: 0.00 b: 0.09 c: 0.15 d: 0.61 e: 0.78

Suppose the randomly selected dorm room has TV. Given this information, what is the conditional probability that the dorm has no refrigerator?

a: 0.00 b: 0.060 c: 0.120 d: 0.170 e: 0.349

The events that "a dorm room has a TV" and "a dorm room has a refrigerator" are:
a)not independent, because P(Refrigerator and TV) ≠ 0.
b)independent, because P(Refrigerator and TV) = P(Refrigerator) ×P(TV).
c)disjoint, because P(Refrigerator and TV) ≠ 0.
d)disjoint, because P(Refrigerator and TV) ≠ P(Refrigerator) ×P(TV).
e)not independent, because P(Refrigerator and TV) ≠ P(Refrigerator) ×P(TV).

Answer 1

Let X be the event that a dorm has refrigerator.

and Y be the event that a dorm has a TV.

Given P(X) = 0.48, P(Y) = 0.43 and P(X U Y) = 0.63

What is the probability that a randomly selected dorm room has both a TV and a refrigerator?

from the formula

0.48 + 0.43 - 0.63 = 0.28 => Probability that a dorm has both refrigerator and TV.

What is the probability that a randomly selected dorm room has either a TV or a refrigerator, but not both?

It is simply P(a dorm has atleast a TV or refrigerator) - P(dorm has both TV and refrigerator)

= 0.63 - 0.28 = 0.35

What is the probability that a randomly selected dorm room has only TV but no refrigerator?

It is given by P(dorm has TV) - P(dorm has TV and refrigerator both)

= 0.43 - 0.28 = 0.15

Suppose the randomly selected dorm room has TV. Given this information, what is the conditional probability that the dorm has no refrigerator?

Let X' be event that a dorm has no refrigerator,

P(dorm has no refrigerator given dorm has TV) = P(X' / Y) = = 0.15 / 0.43 = 0.349

The events that "a dorm room has a TV" and "a dorm room has a refrigerator" are:

It is asked about Y and X.

If X and Y are independent then

P(X)*P(Y) = 0.43*0.48 = 0.20 0.28 - option (e)

Hence X and Y are not independent.

Since 0 Hence X and Y are not disjoint.

We only have option (e) correct in this case.