MA22S6 Homework 2 Solutions
ˆ Question 1
Simultaneously means without replacement, hence we find
4
52
3
2
1
× 51
× 50
× 49
=
1
270725 .
ˆ Question 2
Binary consists of two possible numbers, 0 and 1. On each finger we have two possible
representations (i.e., a 0 or a 1), and we have ten fingers, meaning we have 210 = 1024
possible combinations we can make. If the first number we can make is a 0 then the
largest we can make is 1023.
ˆ Question 3
In the first part of this our answer is simply 10! = 3628800. In the next part, each couple
is considered as an individual unit, which can be ordered in 5! = 120 ways. However
each couple can sit two ways (man first and the woman second or woman first and the
man second). Each couple can sit in 2! = 2 ways, and we have five couples, meaning the
inter-coupling orderings have (2!)5 = 32 combinations, giving us a solution for part (b) of
5!×(2!)5 = 3840 orderings. Lastly, men and women must sit seperately, e.g., the row could
be W M W M W M W M W M . Men can sit in 5! = 120 orderings, as can women. The first
person in the row can be a man or a woman, leading to a solution of 5! × 5! × 2 = 28800.
ˆ Question 4
The Monty Hall problem is controversial and counter-intuitive. We use Bayes’ Theorem
to justify its solutions. Denote by A, B, C and D the event of the prize being behind
doors A, B, C and D respectively, and by O the event of Monty opening door B. We have,
a priori, P(A) = P(B) = PC = P(D) = 14 , i.e., having seen nothing all doors are equally
likely to contain the prize. Assume, without loss of generality, we choose door A, giving
– P(O∣A) = 13 , i.e., if A contains the prize then Monty can open B, C or D with equal
likelihood.
– P(O∣B) = 0, i.e., if B contains the prize then Monty can’t open it.
– P(O∣C) = 21 , i.e., if C contains the prize then Monty can only open B or D .
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– P(O∣D) = 12 , i.e., if D contains the prize then Monty can only open B or C
By the Law of Total Probability
P(O) = P(O∣A)P(A) + P(O∣B)P(B) + P(O∣C)P (C) + P(O∣D)P (D)
=
1 1 1 1 1 1
1 1
× +0× + × + × = .
3 4
4 2 4 2 4 3
Now we consider the probability of success if we don’t switch, and if we do switch, using
Bayes.
P(C∣O) =
P(O∣C)P(C)
P(O)
1
1
3
2 × 4
=
= .
1
8
3
P(O∣D)P(D)
P(D∣O) =
P(O)
1
1
×
3
= 2 1 4 = .
8
3
We see our probability of winning increases from
1
4
when we switch.
ˆ Question 5
Here the number of trucks which blow out, X, is Binomially distributed with probability
0.25 of blowing out per trial, i.e., X ∼ Bin(15.0.25). The probability of k of the 15 trucks
)(0.25)k (0.75)15−k . The probability of three trucks blowing
blowing out is P(X = k) = (15
k
out, P(X = 3) = 0.225. The probability of between three and sixes trucks blowing out is
P(3 ≤ X ≤ 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6), giving 0.225 + 0.225 + 0.165 +
0.09 = 0.705. To find the probability of less than four trucks blowing out we find P(X <
4) = P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3), giving 0.013+0.068+0.156+0.225 = 0.462.
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