Assignment 2 Solutions
Problems
2.43
2.44
2.67
2.71
2.74
2.75
2.86
2.92
Score
6
12
18
10
12
18
12
12
2.43
The number of ways = 𝐶39 · 𝐶56 · 𝐶11 =
9!
6!×3!
×
6!
1!×5!
×
1!
0!×1!
= 504. Number of ways taxis distributed to
airport A is 𝐶39 , number of ways taxis distributed to airport B is 𝐶56 and number of ways taxis
distributed to airport C is 𝐶11 .
2.44
a.
The numerator will equal 𝐶38 · 𝐶55 · 𝐶11 =
8!
5!×3!
×
5!
0!×5!
×
1!
0!×1!
= 56. Since we know that one car has
already been distributed to airport C, there are 8 left over. Number of ways taxis distributed to airport A
is 𝐶38 and number of ways taxis distributed to airport B is 𝐶55 . Thus, P(The car in need of repair being
sent to airport C) = 56/504 = 1/9.
b.
The numerator will equal the number of ways to distribute the three taxis in need of repair to the three
airports with each of them receiving one multiplied by the numbers of ways to distribute the remaining
taxis. This will equal (
3!
1!×1!×1!
6
6!
1
4!×2!
)( 𝐶26 · 𝐶44 ) = ( ×
×
4!
0!×4!
) = 6 × 15 = 90. Airport A is the
Combination of 6 choose 2 since we know that three cars have already been distributed and one is sent to
Airport A and airport B is the Combination of 4 choose 4. Thus, P(Every airport receiving one taxi in
need of repair) = 90/504 = 5/28
2.74
a.
P(A) = .61; P(D) = .3; P(𝐴 ∩ 𝐷) = .2. P(A)P(D) = .183. Thus, P(𝐴 ∩ 𝐷) does not equal P(A)P(D), A
and D are dependent events.
b.
P(B) = .3; P(D) = .3; P(𝐵 ∩ 𝐷) = .09. P(B)P(D) = .09. Thus, P(𝐵 ∩ 𝐷) = P(B)P(D), B and D are
independent events.
c.
P(C) = .09; P(D) = .3; P(𝐶 ∩ 𝐷) = .01. P(C)P(D) = .027. Thus, P(𝐶 ∩ 𝐷) does not equal P(C)P(D), C
and D are dependent events.
Comment on 2.75: Si means first card drawn is a spade.
2.86
a.
No, because P(𝐴 ∪ 𝐵) ≤ 1 and P(𝐴 ∪ 𝐵) = P(A) + P(B) – P(𝐴 ∩ 𝐵). If P(A) = .8 and P(B) = .7 and
P(𝐴 ∩ 𝐵) = .1, than P(𝐴 ∪ 𝐵) = .8 + .7 - .1 = 1.4. 1.4 > 1, which does not satisfy the prior statement that
P(𝐴 ∪ 𝐵) ≤ 1.
b.
The smallest possible value of P(𝐴 ∩ 𝐵) is .5. This is because P(𝐴 ∪ 𝐵) ≤ 1 and P(𝐴 ∪ 𝐵) = P(A) +
P(B) – P(𝐴 ∩ 𝐵). If P(A) = .8 and P(B) = .7, than the smallest value that makes P(𝐴 ∪ 𝐵) ≤ 1 is .5.
c.
No, P(𝐴 ∩ 𝐵) is the intersection of A and B, which means that it would not be possible for P(𝐴 ∩ 𝐵) to
be greater than P(B) which is .7.
d.
The largest possible value of P(𝐴 ∩ 𝐵) is the smaller one of P(A) and P(B) which is .7.