MATH 121 DISCUSSION WEEK 10 SOLUTIONS
1)
(a) Let A = draw even # from first pile, B = draw odd # from second pile, C = draw # less
than 3 from third pile. A =
{
2, 4, 6, 8
}
, B =
{
1, 3, 5, 7
}
, C =
{
1, 2
}
, so P(A) =
4
8
, P(B) =
4
8
, P(C) =
2
8
.
P(A
∩
B
∩
C) = P(A)
×
P(B)
×
P(C) = (
4
8
)(
4
8
)(
2
8
) = (
1
2
)(
1
2
)(
1
4
) =
1
16
.
(b) There are 8
×
8
×
8 = 512 possible combinations of three cards. There are 8 ways you can
draw the same # from each pile:
{
1, 1, 1
}
,
{
2, 2, 2
}
,
{
3, 3, 3
}
,
{
4, 4, 4
}
,
{
5, 5, 5
}
,
{
6, 6, 6
}
,
{
7, 7, 7
}
,
{
8, 8, 8
}
. Thus, P(draw all the same #s) =
8
512
=
1
64
.
(c) P(not all the same #s) = 1  P(all the same #s) = 1 
8
512
=
512
512

8
512
=
504
512
.
2)
(a) There are 365
×
365
×
365
×
365 = 365
4
possible combinations of 4 birthdays. There are 365
ways that all four friends can have the same birthday (i.e. they all have a birthday on Jan.
1, they all have a birthday on Jan. 2, etc., all the way through Dec. 31).
Thus, P(all the same birthday) =
365
365
4
=
1
365
3
=
1
48
,
627
,
125
.
(b) There are 365
×
364
×
363
×
362 combinations of 4 birthdays where each day is different.
Thus, P(no one shares a birthday) =
365
×
364
×
363
×
362
365
4
≈
0
.
9836.
(c) The probability that none of the friends share a birthday was found in part (b). The comple
ment of no one sharing a birthday is that at least 2 of them have a common birthday. Thus,
P(at least 2 share a birthday) = 1  P(no one shares a birthday)
= 1

365
×
364
×
363
×
362
365
4
≈
0
.
0164.
3)
Let A = pick white pair, B = pick blue pair, C = pick black pair, and D = pick yellow pair,
and “1” and “2” denote the first and second pick respectively.
(a) P(B2

A1) =
2
7
(b) P(A2

D1) =
3
7
(c) P(B2

B1) =
1
7
(d) P(D2
∩
B1) = P(D2

B1)
×
P(B1) = P(B1)
×
P(D2

B1) = (
2
8
)(
1
7
) =
2
56
=
1
28
.
(e) P(B2
∩
B1) = P(B2

B1)
×
P(B1) = P(B1)
×
P(B2

B1) = (
2
8
)(
1
7
) =
2
56
=
1
28
.
(f) P(A2
∩
A1) = P(A2

A1)
×
P(A1) = P(A1)
×
P(A2

A1) = (
3
7
)(
2
6
) =
6
42
=
1
7
.
4)
For all four parts, let A = animal is a cat, B = animal is black.
(a) P(A) = P(A
∩
B) + P(A
∩
B
0
) = 5% + 35% = 40%.
(b) P(A) = P(A
∩
B) + P(A
∩
B
0
), so P(A
∩
B
0
) = P(A)  P(A
∩
B) = 50%  2% = 48%.
(c) In this case, we already know that the animal is black, so event B has already occurred. So
we want to know what the probability of A is
given
that B occurred.
P(A

B) =
P
(
A
∩
B
)
P
(
B
)
=
0
.
03
0
.
30
=
1
10
, or 10%.
(d) In this case, we want to know the probability that an animal is black
given
that it is a cat
(i.e. A has already occurred, and we want probability of B given that A occurred).
P(A
0
) = 70%, or 0.70. So P(A) = 1  P(A
0
) = 1  0.70 = 0.30. P(B

A) =
P
(
B
∩
A
)
P
(
A
)
=
0
.
10
0
.
30
=
1
3
.
1