Maths(四)---Probability and Statistics(5)

Exhaustive Events

例一.

Two fair coins are tossed. A is the event that at least one tail is obtained

(a) Describe an event B such that A and B are exhaustive events only.

(b) Describe an event C such that A and C are both mutually exclusive andexhaustive.

Solution

(a)The possibility space S = {HH, HT, TH, TT} and the event A = {HT, TH, TT}.Let B be the event that at least one head is obtained, then B = {HH, HT, TH}.Since A ∪ B = {HH, HT, TH, TT} = S, A and B are exhaustive events.

(b)Let C be the event that no tail is obtained, then C = {HH}. Since A ∪ C ={HH, HT, TH, TT} = S and A ∩ C = φ, A and C are both mutuallyexclusive and exhaustive.

例二.

A and B are two events such that P( A ) =2/3 , P( B ) =2/5 and P( A ∩ B ) =1/15 .Are A and B exhaustive events?

Solution

2/3+2/5-1/15=1,A and B areexhaustive.

例三.

In a class of 40 students all study at least one of the subjects computer science anddiscrete mathematics. 27 attend the computer science class and 32 attend thediscrete mathematics class. Find the probability that a student chosen at randomstudies both computer science and discrete mathematics

Solution

40 = 27 + 32 − n( C ∩ M )

n( C ∩ M ) = 19

Therefore the probability that a student chosen at random studies both computer science and discrete mathematics is P( C ∩ M ) =19/40 .

Random Variables

例一.

Suppose 3 fair coins are tossed. The sample space of this experiment can be written as

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Let the random variable X be the number of tails facing upwards. Then

X(HHH) = 0

X(HHT) = X(HTH) = X(THH) = 1

X(HTT) = X(THT) = X(TTH) = 2

X(TTT) = 3

The range space R = {0, 1, 2, 3}.

Probability Distributions

P(X = x) is often written as p(x).

p(0) = P(X = 0) = P(HHH) =1/8

p(1) = P(X = 1) = P(HHT, HTH, THH) =3/8

p(2) = P(X = 2) = P(HTT, THT, TTH) =3/8

p(3) = P(X = 3) = P(TTT)=1/8

例二.

Suppose two fair dice are tossed. Let the random variable X be the total score onthe two dices. The range space isR = {2, 3, 4, …, 12}.

The probability distribution of X is

例三.

A salesman visits 3 customers A, B and C to sell a product. The probabilities thatA, B and C will order the product are 0.2, 0.3 and 0.5 respectively. Let X be thenumber of customers that will order the product. Find the probability distributionof X.

Solution:

Let A, B and C be the events that customer A, B and C will order the product.

P(X = 0) = P(-A-B-C ) = (1 – 0.2)(1 – 0.3)(1 – 0.5) = 0.28

P(X = 1) = P(A-B-C or -AB-C or -A-BC)= (0.2)(0.7)(0.5) + (0.8)(0.3)(0.5) + (0.8)(0.7)(0.5)= 0.47

P(X = 2) = P(AB-C or A-BC or -ABC)= (0.2)(0.3)(0.5) + (0.2)(0.7)(0.5) + (0.8)(0.3)(0.5)= 0.22

P(X = 3) = P(ABC) = (0.2)(0.3)(0.5) = 0.03

例四.

In a test paper, there are five true-or-false questions. Two marks are awarded foreach correct answer and one mark is deducted for each wrong answer. Suppose astudent answers each question by choosing T or F randomly and let X be the totalmarks he gets.

Solution

The sample space can be wriiten asS = {WWWWW, WWWWC, WWWCW, …, CCCCC}where, for example, the outcome WWWWC means that the first 4 answers arewrong and the fifth answer is correct

Number of outcomes in S = 25 = 32Number of outcomes with k C’s = 5Ck , with k = 0, 1, 2, 3, 4, 5Obviously X = 2k – (5 – k) where k is the number of correct answers. The rangespace of X is R = {-5, -2, 1, 4, 7, 10}.