Download AK/AS/SC MATH 2030.03 Quiz #1, version AS,T,A,T,I,S,T,I,C,S

AK/AS/SC MATH 2030.03
Quiz #1, version A
October 03, 2003
SOLUTIONS
1. (6 points) The letters of the word STATISTICS are arranged in a random order.
Find the probability
(a) they spell statistics.
Answer:
The outcome space
Ω = { all the distinct arrangements of the letters S,T,A,T,I,S,T,I,C,S }, where
S appears 3 times, T appears 3 times, I appears twice, the other two letters - once
each.
So, the number of all the distinct arrangements is equal
10!
#(Ω) =
= 50 400.
3!3!2!
There is only one arrangement when the letters S,T,A,T,I,S,T,I,C,S spell
statistics. So, #(A) = 1.
#(A)
1
Hence, P (A) =
=
.
#(Ω)
50 400
(b) the same letter occurs at each end.
Answer:
The outcome space is the same as in part (a).
B = { all the distinct arrangements of the forms S
S
or T
T or I
I }.
So, by the rule of sum
8!
8!
8!
#(B) =
+
+
= 7 840.
3!2! 3!2! 3!3!
7 840
7
#(B)
Hence, P (B) =
=
= .
#(Ω)
50 400
45
2. (6 points) Suppose that A, B and C are three events such that A and B are mutually
exclusive, C is independent of both A and B. Also, P (A) = 0.3, P (B) = 0.5, P (C) =
0.4. Calculate:
S
(a) P (A B)
Answer:
S
P (A B) = P (A) + P (B) − P (AB) = 0.3 + 0.5 − 0 = 0.8.
(b) P (AC)
Answer:
P (AC) = P (A)P (C) = 0.3 · 0.4 = 0.12.
S
(c) P (A C).
Answer:
S
P (A C) = P (A) + P (C) − P (AC) = 0.3 + 0.4 − 0.12 = 0.58.
3. (8 points) There are five bowls. Each of the first two bowls (type A1 ) contains two
white and three black balls, each of the other two bowls (type A2 ) contains one white
and four black balls, and the fifth bowl (type A3 ) contains four white and one black
ball. We select a bowl at random and draw a ball from the bowl at random.
(a) Find the probability that the drawn ball is white.
Answer:
By the rule of average conditional probabilities,
P (white) = P (white | A1 )P (A1 ) + P (white | A2 )P (A2 ) + P (white | A3 )P (A3 )
2 2 1 2 4 1
2
= · + · + · = .
5 5 5 5 5 5
5
Equivalently, the probability P (white) can be calculated using a tree diagram.
(b) Given that the drawn ball is black, what is the probability that it came from the
fifth bowl?
Answer:
By Bayes’ Rule,
P (black | A3 )P (A3 )
P (A3 | black) =
P (black | A1 )P (A1 ) + P (black | A2 )P (A2 ) + P (black | A3 )P (A3 )
1 1
·
1
= 3 2 54 52 1 1 = .
15
· +5·5+5·5
5 5
The end.
2