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Statistics 116 - Fall 2004
Theory of Probability
Practice Midterm # 2 – Solutions
Q. 1) A sample of 3 items is selected at random from a box containing 20 items
of which 4 are defective. Find the expected number of defective items in
the sample.
Solution: Let X be the number of defective items, them the p.m.f. of X
is
16 4
pX (j) =
j
3−j
20
3
for 0 ≤ j ≤ 3 an integer. The expected number of defective items is
therefore
16 4
3
X
j 3−j
j×
20
3
j=0
=1×
4
1
16
2
20
3
+2×
4
2
16
1
20
3
+3×
4
3
16
0
20
3
.
Q. 2) A box contains 7 red and 8 blue marbles. Two marbles are withdrawn
randomly. If they are the same color, then you win $1.10; if they are
different colors, then you win $-1.00 (that is, you lose $1.00).
Compute
(a) the expected value of the amount you win;
(b) the variance of the amount you win.
Solution: If X represents the winnings, then
X({R, B}) = −1.00, X({R, R}) = X({B, B}) = 1.10.
1
The p.m.f. of X is therefore
7
1
pX (−1) =
=
8
15
2
1
8
15 7 8
0
pX (1.10) =
2 +
15
2
=1−
=
7
2
8
0
15
2
8
15
7
.
15
Therefore, the expected winnings are
E(X) = −1 ∗
8
7
+ 1.1 ∗ .
15
15
Q. 3) Suppose that each child born to a couple is a boy with probability 0.48
and a girl with probability 0.52, independent of the sex distribution of the
other children in the family. For a couple having 5 children, compute the
probabilities of the following events:
(a) All children are of the same sex.
(b) The 3 eldest are boys and the others girls.
Solution:
(a) The event that all children are the same sex is the event that they are
all boys, which has probability (0.48)5 union the event that they are
all girls, which has probability (0.52)5 . The probability is therefore
(0.48)5 + (0.52)5 .
(b) The event that the three eldest are boys and the others girls is
(0.48)3 · (0.52)2 .
Q. 4) In Figure 1, each of the switches are closed independently with probability
pi . Find
(a) the probability that current flows from A to B
(b) the probability that both switches 1 and 2 are closed, given that
current flows from A to B.
Solution:
2
Figure 1: Circuit # 1 - 4 (a)
(a) If Ei is the event that the i-th switch is closed, and A → B is the
event that current flows from A to B, then
A → B = ((E1 ∩ E2 ) ∪ (E3 ∩ E4 )) ∩ E5 .
As all switches are independent,
P (A → B) = P (((E1 ∩ E2 ) ∪ (E3 ∩ E4 )) ∩ E5 )
= p5 · P (((E1 ∩ E2 ) ∪ (E3 ∩ E4 )))
= p5 (p1 p2 + p3 p4 − p1 p2 p3 p4 ).
(b) We want to compute
P (E1 ∩ E2 ∩ ((E1 ∩ E2 ) ∪ (E3 ∩ E4 )) ∩ E5 )
P (A → B)
P (E1 ∩ E2 ∩ E5 ∪ E1 ∩ E2 ∩ E3 ∩ E4 ∩ E5 )
=
p5 (p1 p2 + p3 p4 − p1 p2 p3 p4 )
p 1 p2 p5 + p 1 p2 p3 p4 p 5 − p1 p2 p 3 p 4 p5
=
p5 (p1 p2 + p3 p4 − p1 p2 p3 p4 )
p 1 p2
.
=
p 1 p 2 + p 3 p 4 − p1 p 2 p 3 p4
P (E1 ∩ E2 |A → B) =
Q. 5) You are to draw 5 balls with replacement from an urn containing 30 balls,
numbered 1 to 30. What is
(a) the p.m.f. of the random variable X = largest number of the 5 balls drawn;
(b) the probability of X being greater or equal to 25?
Solution:
(a) The event {X = i} is the event that we draw 4 balls with value less
than or equal to i and at least 1 ball of value i. It is actually easier
to compute the probability that {X ≤ i} which is
P (X ≤ i) =
3
i 5
.
30
Now
P (X = i) = P (X ≤ i) − P (X ≤ i − 1) =
i5 − (i − 1)5
.
305
(b)
P (X ≥ 25) = 1 − P (X < 25)
= 1 − P (X ≤ 24)
4
=1−
24
30
5
.