THEORY OF PROBABILITY: HOMEWORK 4 SOLUTIONS
VLADIMIR KOBZAR
All Problems and Theoretical Exercises are from Ross, Chapter 2.
Problem (45).
(a) In the first k ´ 1 tries, there are k ´ 1 choices of
the n´1, n´2, ..., k keys that do not work from n, n´1, ..., n´k
keys respectively. On the kth try, 1 choice works from n ´ k ` 1
keys.
1
pn ´ 1qpn ´ 2q...pn ´ k ` 1q ¨ 1
“
npn ´ 1qpn ´ 2q...pn ´ kqpn ´ k ` 1q
n
.
(b) In the first k ´ 1 tries, there are k ´ 1 choices from n ´ 1 keys
that do not work and on the kth try 1 choice works, in each
case, from n keys.
pn ´ 1qk´1
nk
Problem (52).
(a) There are 8 choices from the 20, 18, 16, ..., 6 shoes
from 20, 19, 18, ..., 13 respectively.
20 ¨ 18 ¨ 16 ¨ ... ¨ 6
20 ¨ 19 ¨ ... ¨ 13
(b) Choose the complete pair from 10, choose 6 incomplete pairs
from 9, and choose the left or right shoe for each incomplete
pair.
`˘
10 96 26
`20˘
8
Problem (54). Let Ei denote the event that the bridge hand is void
in the ith suit, for 1 ď i ď 4. By the exclusion-inclusion identity,
Date: July 16, 2016.
1
2
VLADIMIR KOBZAR
`39˘
˘
P pEi q “ `13
52
13
`26˘
˘
P pEi X Ej q “ `13
52
13
`13˘
˘
P pEi X Ej X Ek q “ `13
52
13
4
ď
ˆ ˙
ˆ ˙
4
4
P p Ei q “ 4P pEi q ´
P pEi X Ej q `
P pEi X Ej X Ek q “
2
3
i“1
` ˘
`26˘
4 39
´
6
`4
“ 13 `52˘13
13
Problem (Theoretical Exercise 18). If we start with a head, then the
second outcome must be a tail and the remaining outcomes represent
the number of ways of tossing a coin n ´ 2 times such that successive
heads never appear, i.e. fn´2 . If we start with a tail, then the remaining outcomes represent the number of ways of tossing a coin n ´ 1
times, such that successive heads never appear, i.e. fn´1 . Since the
first outcome may be either a head or a tail, fn “ fn´1 ` fn´2 . Since
the total outcomes are 2n , Pn “ 2fnn .
f2 “ 3, f3 “ 5, ...f10 “ 144. P10 “
144
210
References
[1] Ross, A First Course in Probability (9th ed., 2014)