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

the correct option is 2.

step-by-step explanation:

according to the given information triangle abc is a right angle triangle with right angle at a. segment ab is 5 and segment ac is 7. point d is on segment bc and angles adb and adc are right angles.

it is given that triangles abd, cad, and cba are similar.

two triangles are similar if their corresponding sides are proportional.

in triangle cba, using pythagoras theorem

                      )

triangle abd and cba are similar,

                            (2)

triangle cad and cba are similar,

                      )

using (1), (2) and (3), we get

hence proved.

therefore option 2 is correct.