1. No category

Summary of Lecture 3 Counting Methods The multiplication rule can

Summary of Lecture 3
Counting Methods
The multiplication rule can be stated as, β€œIf there are n1 outcomes in experiment
1, n2 outcomes in experiment 2, and so on up to nk outcomes for experiment k,
then there are n1 x n2 x … x nk outcomes if the k experiments are done in series”.
The multiplicative rule can be applied to calculating the number of ways of
arranging n objects which is,
π‘π‘œ. π‘œπ‘“ π‘€π‘Žπ‘¦π‘  π‘œπ‘“ π‘Žπ‘Ÿπ‘Ÿπ‘Žπ‘›π‘”π‘–π‘›π‘” 𝑛 π‘œπ‘π‘—π‘’π‘π‘‘π‘  = 𝑛!
Now, if we want to compute the number of ways of selecting k objects from a
total of n where the order of selection is important, the calculation is made by
permutations.
𝑛
π‘˜π‘ƒ
=
𝑛!
(𝑛 βˆ’ π‘˜)!
The situation is slightly different if the order of selection is not important. That is,
any group of k objects, whatever the order in which they are selected, are
equivalent. If the order is not important, the calculation is made by
combinations
𝑛
π‘˜πΆ
=
𝑛!
(𝑛 βˆ’ π‘˜)! π‘˜!
E 243 Recitation 2 Spring 2015
February 6, 2015
Examples
1) A multiple choice test consists of 5 questions, each having 4 possible
answers of which only one is correct.
a. In how many ways can a student answer the questions?
b. In how many ways can a student answer all 5 questions
incorrectly?
c. What is the probability that a random selection of answers will
result in the student getting at least one correct answer?
2) What is the probability of at least one shared birthday in a group of 30
people?
3) There are ten socks in a bag, 4 of which are red and 6 are black. If you pick
two socks from the bag, what is the probability that they will match?
Practice Problems
4) There are 9 candidates to fill 3 positions on a faculty committee of whom 2
are assistant professors, 2 are associate professors and 5 are full
professors. If the selection of the three new committee members is
completely random,
a. What is the number of ways that there will be one assistant
professor and two full professors chosen?
b. What is the probability that all three new members will be full
professors?
5) Two officers, a president and a treasurer, are to be chosen from the 50
members of the Stevens' yacht club. How many different choices are there
if:
a. There are no restrictions.
b. David Smith will serve as an officer only if he is the president.
Recitation 2 Solutions
Problem 1
a) No. of ways answering question 1 =4
no. of ways answering question 2= 4
…
no. of ways answering question 5 = 4
Using the multiplication rule, no. of ways answering all questions = 45 = 1024
b) no. of ways answering each question incorrectly = 3
no. of questions = 5
Using multiplication rule, no. of ways of answering all questions incorrectly = 35 =
243
c) P (at least one Correct answer) = 1 – P (no. Correct answers)=1 βˆ’
243
1024
= 0.763
Problem 2
Let π‘₯= (no. of shared birthdays)
P (π‘₯ is at least 1) = P (X β‰₯ 1) = 1- P (X=0)
Now, P (π‘₯ = 0) i.e. probability of no shared birthdays is given by
π‘›π‘œ. π‘œπ‘“ π‘€π‘Žπ‘¦π‘  𝑖𝑛 π‘€β„Žπ‘–π‘β„Ž π‘‘β„Žπ‘’π‘ π‘’ 30 π‘π‘’π‘œπ‘π‘™π‘’ π‘π‘Žπ‘›π‘›π‘œπ‘‘ π‘ β„Žπ‘Žπ‘Ÿπ‘’ π‘Ž π‘π‘–π‘Ÿπ‘‘β„Žπ‘‘π‘Žπ‘¦
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘œ. π‘œπ‘“ π‘€π‘Žπ‘¦π‘  30 π‘π‘’π‘œπ‘π‘™π‘’ π‘π‘Žπ‘› β„Žπ‘Žπ‘£π‘’ π‘π‘–π‘Ÿπ‘‘β„Žπ‘‘π‘Žπ‘¦π‘ 
Total no. of ways 30 people can have birthdays = 365×365×… ×(30 times) = 36530
No. of ways 30 people can have birthdays without sharing
= 365 × 364 × 363 … × (365 βˆ’ 30 + 1) =
(π‘₯ β‰₯ 1) = 1 βˆ’
365!
335!
365!
= 1 βˆ’ 0.294 = 0.706
335! 36530
70.6 % chance of a shared birthday
Problem 3
P (2 socks will match) =
π‘›π‘œ.π‘œπ‘“ π‘€π‘Žπ‘¦π‘  π‘œπ‘“ 𝑠𝑒𝑙𝑒𝑐𝑑𝑖𝑛𝑔 π‘π‘Žπ‘–π‘Ÿπ‘’π‘‘ π‘ π‘œπ‘π‘˜π‘ 
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘œ.π‘œπ‘“ π‘€π‘Žπ‘¦π‘  π‘œπ‘“ 𝑠𝑒𝑙𝑒𝑐𝑑𝑖𝑛𝑔 2 π‘ π‘œπ‘π‘˜π‘  π‘“π‘Ÿπ‘œπ‘š 10
6+15
45
=
4𝐢 6𝐢+ 6𝐢 4𝐢
2 0
2 0
10𝐢
2
= 0.467
Problem 4
2 Assistants, 2 Associates, 5 Full
a) No. of ways 1 assistant and 2 full professors are chosen
2
1𝐢
× 52𝐢 = 2 × 10 = 20
b) P(all members are full professors) = =
5 4
π‘›π‘œ.π‘œπ‘“ π‘€π‘Žπ‘¦π‘  π‘œπ‘“ π‘β„Žπ‘œπ‘ π‘–π‘›π‘” 3 𝑓𝑒𝑙𝑙 π‘π‘Ÿπ‘œπ‘“π‘’π‘ π‘ π‘œπ‘Ÿπ‘ 
3𝐢 0𝐢 = 0.119
=
9
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘œ.π‘œπ‘“ π‘€π‘Žπ‘¦π‘  π‘œπ‘“ π‘β„Žπ‘œπ‘ π‘–π‘›π‘” 3 π‘“π‘Ÿπ‘œπ‘š 9
3𝐢
Problem 5
a) No. of ways of choosing the president = 50
No. of ways of choosing the treasurer (after choosing the president) = 49
Using the multiplication rule = 50×49 = 2450 (or use permutation)
π·π‘Žπ‘£π‘–π‘‘ π‘†π‘šπ‘–π‘‘β„Ž 𝑀𝑖𝑙𝑙 π‘ π‘’π‘Ÿπ‘£π‘’ π‘Žπ‘  π‘π‘Ÿπ‘’π‘ π‘–π‘‘π‘’π‘›π‘‘
π‘œπ‘Ÿ
b) {
β‡’
π·π‘Žπ‘£π‘–π‘‘ π‘†π‘šπ‘–π‘‘β„Ž 𝑀𝑖𝑙𝑙 𝑁𝑂𝑇 π‘ π‘’π‘Ÿπ‘£π‘’ π‘Žπ‘  π‘π‘Ÿπ‘’π‘ π‘–π‘‘π‘’π‘›π‘‘
π‘›π‘œ. π‘œπ‘“ π‘€π‘Žπ‘¦π‘  π‘œπ‘“ π‘β„Žπ‘œπ‘œπ‘ π‘–π‘›π‘” (π‘π‘Ÿπ‘’π‘ π‘–π‘‘π‘’π‘›π‘‘ × π‘‘π‘Ÿπ‘’π‘Žπ‘ π‘’π‘Ÿπ‘’π‘Ÿ)
+
{
β‡’
π‘›π‘œ. π‘œπ‘“ π‘€π‘Žπ‘¦π‘  π‘œπ‘“ π‘β„Žπ‘œπ‘œπ‘ π‘–π‘›π‘” (π‘π‘Ÿπ‘’π‘ π‘–π‘‘π‘’π‘›π‘‘ × π‘‘π‘Ÿπ‘’π‘ π‘Žπ‘ π‘’π‘Ÿπ‘’π‘Ÿ)
1 × 49
{ +
= 2401
49 × 48
=

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