Download RQ 6 - Solutions

UCSC
AMS/ECON 11B
FALL 2007
Review Questions 6
Basic Probability
1. A box contains 2 red tickets and 3 blue tickets. Four tickets are drawn from the box with
replacement.
a. What is the probability that exactly two of the four tickets will be blue?
2 2 2
= 0.3456.
P (exactly two blue tickets) = 4 C2 35
5
b. What is the probability that at least three of the four tickets will be red?
P (at least three red tickets) = P (exactly three red tickets) + P (exactly four red tickets)
3 3 2 4
= 0.1792.
= 4 C3 52
5 + 5
c. What is the probability that none of the four tickets will be red?
d. If 20 tickets are selected with replacement, what is the probability that 10 of the tickets
are red?
Comments: When n tickets are drawn with replacement from a box with r red tickets and
m − r blue tickets (so m tickets in all), we assume that each draw is independent of the
others, and use this assumption to compute the probability of seeing exactly k red tickets in
n draws as follows.
i. The probability of observing any specific sequence of draws that results in k red and
r k m−r n−k
n−k blue tickets, is m
. This follows from the assumption of independence
m
because
k R and n−k B
r k m − r n−k
z
}|
{
.
P (RRB . . . BR) = P (R)P (R)P (B) · · · P (B)P (R) =
m
m
In other words, the order of the red and blue tickets in the sequence of draws does not
affect the probability—only the number of red and blue tickets in the sequence does.
ii. There are n Ck different sequences of draws that result in exactly k red and n − k
non-red tickets, and the probability of each is that given in i. This means that the
probability of observing exactly k red tickets and n − k blue tickets in n draws (with
replacement) from the box is
n
Ck terms
z
}|
{
r k m − r n−k
r k m − r n−k
r k m − r n−k
=
+ ··· +
= n Ck ·
.
m
m
m
m
m
m
Also, there is nothing special about drawing colored tickets from a box. The same analysis
applies to any scenario where an experiment with two outcomes is repeated n times, in such
a way that the repetitions are independent of each other. If the two outcomes are A and B,
with P (A) = p and P (B) = 1 − P (A) = 1 − p, then the probability of observing exactly k
A’s and n − k B’s in n independent repetitions of the experiment is
n Ck p
k
(1 − p)n−k .
For example, if you are tossing a (fair) coin n times, then the probability of observing exactly
k H and n − k T is
k n−k
1
1
n Ck
= n .
n Ck
2
2
2
2. A (bigger) box contains 75 red tickets, 25 green tickets and 20 blue tickets. Ten tickets are
drawn without replacement and without attention to the order.
a. What is the probability that exactly six of the selected tickets will be red?
There are 120 tickets total, and 120 C10 different combinations of 10 tickets that can be
drawn (without replacement) from the box. Of these, there are exactly 75 C6 · 45 C4
combinations that contain exactly 6 red tickets, since there are 75 C6 ways to select the
red tickets and 45 C4 ways to select the non-red ones. Therefore
P (exactly 6 of 10 tickets drawn are red) =
· 45 C4
≈ 0.2585.
120 C10
75 C6
b. What is the probability that no more than 3 of the selected tickets will be red?
No more than 3 red tickets means: exactly 3 red tickets; or exactly 2 red tickets; or
exactly 1 red ticket; or no red tickets. These events are mutually exclusive in pairs, so
the probability of no more than 3 red tickets is the sum of the probabilities of these four
events. These probabilities are computed in the same way we did a.:
45 C10
• P (no red tickets) = P (all non-red tickets) =
≈ 0.0000275;
120 C10
75 C1 45 C9
• P (exactly one red ticket) =
≈ 0.0005726;
120 C10
75 C2 45 C98
≈ 0.0051535;
• P (exactly two red tickets) =
120 C10
75 C3 45 C7
• P (exactly three red tickets) =
≈ 0.0264005;
120 C10
Thus,
P (no more than 3 red tickets) =
45 C10
+
75 C1 45 C9
+
75 C2 45 C98
+
75 C3 45 C7
120 C10
≈ 0.0321541.
c. What is the probability that there will be exactly six red tickets and at least two blue
tickets in the sample?
Exactly 6 red tickets and at least 2 blue tickets means; 6 red, 2 blue and 2 green; or 6 red,
3 blue and 1 green; or 6 red and 4 blue (no green). Once again, these three possibilities
are mutually exclusive, so the probability that we seek is the sum of the probabilities of
these three events.
• P (6 red, 2 blue and 2 green) =
• P (6 red, 3 blue and 1 green) =
• P (6 red and 4 blue) =
75 C6 20 C2 25 C2
120 C10
75 C6 20 C3 25 C1
120 C10
75 C6 20 C4
120 C10
≈ 0.0988858.
≈ 0.0494429.
≈ 0.0084053.
Thus,
P (6 red and at least 2 blue) =
75 C6 20 C2 25 C2
+
75 C6 20 C3 25 C1
+
75 C6 20 C4
120 C10
≈ 0.1567399.
3. In the Duke’s ticket chamber, there are 100 boxes—70 black boxes and 30 brown boxes.
Each black box contains 500 red tickets and 100 blue tickets, and in each brown box there
are 40 red tickets and 60 blue tickets. One dark and stormy night, the Duke’s ticket flunky
enters the ticket chamber, which is shrouded in total darkness, selects a box at random and
draws two tickets without replacement from the selected box.
a. What is the probability that both tickets are red?
Denote by RR the event that both tickets are red, by BLK the event that the box was
black and by BRN the event that the box was brown. Then
P (RR) = P (RR ∩ BLK) + P (RR ∩ BRN )
= P (RR|BLK)P (BLK) + P (RR|BRN )P (BRN )
7
3
500 C2
40 C2
=
·
·
+
600 C2 10
100 C2 10
≈ 0.533221.
b. What is the probability that both are blue?
Continuing in the same manner, denote by BB the event that both tickets are blue, then
P (BB) = P (BB ∩ BLK) + P (BB ∩ BRN )
= P (BB|BLK)P (BLK) + P (BB|BRN )P (BRN )
7
3
100 C2
60 C2
·
·
=
+
600 C2 10
100 C2 10
≈ 0.126555.
c. After leaving the ticket chamber, the ticket page sees that she has drawn one ticket of
each color. What is the probability that the tickets came from a black box?
Suppose that RB denotes the event that one ticket of each color is drawn. Then, first
note that P (RB)+P (RR)+P (BB) = 1, so P (RB) = 1−P (RR)−P (BB) ≈ 0.340224.
So,
500·100
7
· 10
P (RB|BLK)P (BLK)
0.19477
600 C2
P (BLK|RB) =
≈
≈
≈ 0.572473.
P (RB)
0.340224
0.340224
4. Given that P (E|F ) = 0.5, P (E|F 0 ) = 0.1 and P (F ) = 0.6 compute
a. P (E ∩ F ) = P (E|F )P (F ) = 0.5 · 0.6 = 0.3
Note: We need to compute P (E ∩ F 0 ) before we compute P (E ∪ F ).
b. P (E ∩ F 0 ) = P (E|F 0 )P (F 0 ) = 0.1 · 0.4 = 0.04.
c. P (E ∪ F ) = P (E) + P (F ) − P (E ∩ F )
z
}|
{
= [P (E ∩ F ) + P (E ∩ F 0 )] + P (F ) − P (E ∩ F ) = P (F ) + P (E ∩ F 0 ) = 0.64.
d. P (F |E) =
0.3
P (E ∩ F )
=
≈ 0.882353.
P (E)
0.34
5. Box A contains 100 red tickets and 20 blue tickets, and box B contains 60 red tickets and
60 blue tickets. A box is selected at random and 10 tickets are drawn without replacement
from the box. Exactly 6 of the tickets in the sample are red. What is the probability that
the tickets came from box A?
Denote by E the event that exactly 6 of 10 sampled tickets are red. Then
P (A|E) =
P (E|A)P (A)
P (E|A)P (A) + P (E|B)P (B)
1
2
120 C10
100 C6 20 C4 1
60 C6 60 C4 1
· +
·
2
2
120 C10
120 C10
100 C6 20 C4
=
≈ 0.191315.
·