Question

If the probabilities are 0.87, 0.36, and 0.29 that a family owns a color television set, a black television set, or both kind of sets:

a.what is the probability that a family owns one or the other or both kind of sets?
b.what is the probability that a family does not own any kind of set?
c.what is the probability that a family owns only one kind of set?

Answer 1

P(A)=family owns a color television set=0.87

P(B)=family owns a black television set=0.36

p(A
`\cap` B)=family owns both kind of television set=0.29

(a) Probability that a family owns one or the other or both kind of sets=P(A
`\cup` B)= P(A) + P(B) - P(A
`\cap` B)

=0.87+0.36-0.29

=1.23-0.29

0.94

(b) Probability that a family does not own any kind of set= 1 - P(A
`\cup` B ) = 1-0.94

=0.06

(c) Probability that a family owns only one kind of set= P(A complement) + P(B complement)= P(A)- P(A
`\cap` B) + P(B)- P(A
`\cap` B)

=0.87-0.29+0.36-0.29

=0.58+0.07

=0.65

Answer 2

Answer : (a) 0.94

(b) 0.06

(c) 0.58 and 0.07

Explanation :

We have given that

Event A : a family owns a color television set

Event B: a family owns a black television set

so,

`P(A) = 0.87`

`P(B) =0.36`

`P(A\cap B)= 0.29`

Now, we have to calculate ,

`(a) \text{P( a family owns one or other or both )}`=
`P(A\cup B)`

And we know that ,

`P(A \cup B)= P(A)+ P(B)- P(A\cap B)`

=
`=0.87+0.36-0.29`

`=0.94`

`\text{(b) P( neither of any kind) }` =
`1- P(A\cup B)`

`=1-0.94`

`=0.06`

(c)
`P( only A)= P(A)- P(A\cap B)`

`=0.87-0.29`

`= 0.58`

`P( only B) = P(B)- P(A\cap B)`

`=0.36-0.29`

`=0.07`