Respuesta :
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,
[tex]P(A) = 0.87[/tex]
[tex]P(B) =0.36[/tex]
[tex]P(A\cap B)= 0.29[/tex]
Now, we have to calculate ,
[tex](a) \text{P( a family owns one or other or both )}[/tex]=[tex]P(A\cup B)[/tex]
And we know that ,
[tex]P(A \cup B)= P(A)+ P(B)- P(A\cap B)[/tex]
=[tex]=0.87+0.36-0.29[/tex]
[tex]=0.94[/tex]
[tex]\text{(b) P( neither of any kind) }[/tex] = [tex]1- P(A\cup B)[/tex]
[tex]=1-0.94[/tex]
[tex]=0.06[/tex]
(c) [tex]P( only A)= P(A)- P(A\cap B)[/tex]
[tex]=0.87-0.29[/tex]
[tex]= 0.58[/tex]
[tex]P( only B) = P(B)- P(A\cap B)[/tex]
[tex]=0.36-0.29[/tex]
[tex]=0.07[/tex]
P(A)=family owns a color television set=0.87
P(B)=family owns a black television set=0.36
p(A [tex]\cap[/tex] 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 [tex]\cup[/tex] B)= P(A) + P(B) - P(A [tex]\cap[/tex] 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 [tex]\cup[/tex] 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 [tex]\cap[/tex] B) + P(B)- P(A [tex]\cap[/tex] B)
=0.87-0.29+0.36-0.29
=0.58+0.07
=0.65
