hw4 - MATH 471 HW 4 Solutions Pengsheng Ji Oce Hours...

MATH 471 HW 4 Solutions
Pengsheng Ji
Office Hours:
4:05-5:05 PM Tuesday
5:20-6:20 PM Thursday
218 Mallot Hall
2.4.6
Let A=”1 was received”, B=”1 was sent”. Since they are received as sent with
probability 0.9 but errors occur with probability 0.1, we have
P ( A | B ) = 0 . 9 , P ( A
c
| B ) = 0 . 1 , P ( A
c
| B
c
) = 0 . 9 , P ( A | B
c
) = 0 . 1 .
Then
P ( A ∩ B ) = P ( A | B ) P ( B ) = 0 . 9 * 0 . 5 = 0 . 45 ,
P ( A ∩ B
c
) = P ( A | B
c
) P ( B
c
) = 0 . 1 * 0 . 5 = 0 . 05 .
Hence, the probability that a 1 was sent given that we received a 1 is given by
P ( B | A ) =
P ( A ∩ B )
P ( A )
=
0 . 45
0 . 45 + 0 . 05
= 0 . 9 .
2.4.17
Let A=”the person voted”, B
1
=”a Conservative”, B
2
=”a Liberal”, B
3
=”an
Independent.” By the Bayes Formula, we have the probability she is Liberal given
that she is a voter is given by
P ( B
2
| A ) =
P ( B
2
) P ( A | B
2
)
∑
3
i =1
P ( B
i
) P ( A | B
i
)
=
0 . 5 * 0 . 8
0 . 3 * 2 / 3 + 0 . 5 * 0 . 8 + 0 . 2 * 0 . 5
= 4 / 7 .
2.6.11
Denote these three events by A,B and C, respectively. Then P(A)=1/4,
P(B)=1/3, P(C)=1/2. Since they are independent, the probability exactly one
will occur is
P ( A ∩ B
c
∩ C
c
) + P ( A
c
∩ B ∩ C
c
) + P ( A
c
∩ B
c
∩ C )
=
1
4
* (1 -
1
3
) * (1 -
1
2
) + (1 -
1
4
) *
1
3
* (1 -
1
2
) + (1 -
1
4
)(1 -
1
3
) *
1
2
=
11
24
.

2.6.23
(a) Let A=”a driver will have an accident”, B
i
, i = 1 , 2 , 3 , 4, be the four groups
from the youngest to the oldest. By the Law of Total Probability, the probability