Probability 1

PROBALILITY THEORY AND APPLICATIONS

PROBABILITY THEORY  Probability

Theory is the study of ‘chance’.  It has vitally important applications in  Sciences  Economics  Politics  Sports  Life insurance  Quality control  Production management and host of other areas.

DEFINITIONS  Experiment:

An act whose outcomes are known in advance but is not possible to predict one with surety.  Sample

Space (S): The set of all possible outcomes of an experiment.

SAMPLE SPACE

SAMPLE SPACE

DEFINITIONS  Event:

Any subset of the sample space.  Equally likely events: Equal chance of happening each of the outcome.  Exhaustive events: The events covering the entire sample space, that is A or B = S.

DEFINITIONS  Exclusive

events: Events that do not have any common happening.  Sure event: An event that is certain to happen.  Impossible event: An event that is not at all possible to happen.

MORE ABOUT..

FORMULAE  For

any event A, P(A) = n (A) / n (S) And 0 ≤ P(A) ≤ 1.  For two events A and B, P(A or B) = P(A) + P(B) − P(A and B)  If A and B are mutually exclusive events, P(A or B) = P(A) + P(B)

EXAMPLE 1  Find

P(A), P(B) and P(A or B).

8 6

13

Total number of outcomes = 50

SOLUTION n

(S) = 50  n (A) = 8 + 6 = 14 ⇒ P(A) = 14/50  n (B) = 13 + 6 = 19 ⇒ P(B) = 19/50  n (A and B) = 6 ⇒ P(A and B) = 6/50  P(A or B) = P(A) + P(B) − P(A and B) = 14/50 + 19/50 − 6/50 = 27/50

EXAMPLE 2 A

fair coin is tossed three times. Find the probability of getting 1 head.  For the experiment, Sample Space =S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} i.e., n (S) = 8

Continued..  Let

Event A: getting 1 head Thus, A = {HTT, THT, TTH} i.e., n (A) = 3 Hence, probability of getting 1 head = P(A) = n (A)/n (S) = 3/8

EXAMPLE 3 A

pair of dice is thrown. Find the probability of getting (i) a total of 10 (ii) both odd digits (iii) a total that is multiple 3

Continued..  For

the experiment, S = {(1, 1), (1, 2), (1, 3),….., (6, 6)} i.e., n (S) = 36 Let event A: total is 10 A = {(4, 6), (5, 5), (6, 4)} ∴ n (A) = 3 Thus, P(A) = 3/36 = 1/12

Continued.. Let event B: both are odd digits B = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)} ∴ n (B) = 9 and P(B) = 9/36 = ¼ Let event C: total is a multiple of 3 C = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)} ∴ n (C) = 12 and P(C) = 12/36 = 1/3

EXAMPLE 4  From

a well-shuffled pack of 52 cards, a card is drawn at random. What is the probability that it is (i) red (ii) spade (iii) picture (iv) face (v) heart or king.

Continued..  For

the experiment, S = {AS, 2S, .., KS, AH, 2H, .., KH, AC, 2C, .., KC, AD, 2D, .., KD} i.e., n (S) = 52 Let event A: red card is drawn (Heart and Diamond) n (A) = 26 and P(A) = 26/52 = 1/2

Continued.. Let event B: spade card is drawn (There are 13 cards of spade) n (B) = 13 and P(B) = 13/52 = ¼ Let event C: picture card is drawn (There are 16 picture cards) n (C) = 16 and P(B) = 16/52 = 4/13

Continued.. Let event D: face card is drawn (There are 12 face cards) n (D) = 12 and P(D) = 12/52 = 3/13 Let event E: Heart or King card is drawn (There are 13 hearts, 4 kings; 1 common) n (E) = 16 and P(E) = 16/52 = 4/13

EXAMPLE 5  Soldier

O'Gara, a prison administrator, has been reviewing the prison records on attempted escapes by inmates. He has data covering last 45 years that the prison has been open, arranged for seasons. The data are summarized in the table:

TABLE Escapes 0 1-5 6-10 11-15 16-20 21-25 25 Total

Winter 3 15 15 5 3 2 2 45

Spring 2 10 12 8 4 4 5 45

Summer 1 11 11 7 6 5 4 45

Fall 0 12 16 7 5 3 2 45

EXAMPLE … 1 What is the probability that in a year selected at random, the number of escapes was between 16-20 during the winter? 2 What is the probability that more than 10 escapes were attempted during a randomly chosen summer season?

SOLUTION 



The probability that in a year selected at random, the number of escapes was between 16-20 during the winter = 3/45 = 1/15. The probability that more than 10 escapes were attempted during a randomly chosen summer season = 22/45.

READY FOR PRACTICE?

READY FOR PRACTICE?

EXERCISE

EXERCISE

EXERCISE

EXERCISE

EXERCISE

FORMULAE  For

exhaustive events A and not A, P(A) + P (not A) = 1  For independent events A and B, P(A and B) = P(A) × P(B)  For dependent events A and B, P(A and B) = P(AB) × P(B) = P(BA) × P(A)

EXAMPLE 1

EXAMPLE 2

Total number of marbles = 5 + 3 + 7 = 15 One marble is selected at a time, ∴ n (S) = 15 Let event A: the marble is red ∴ n (A) = 3 and P(A) = 3/15 = 1/5

EXAMPLE 3

Continued.. Let event B: the marble is green ∴ n (B) = 5 and P(B) = 5/15 = 1/3 Let event C: the marble is blue ∴ n (C) = 7 and P(C) = 7/15 Let event D: the marble is not red (not A) ∴ P(D) = 1 – P(A) = 4/5

Continued.. Let event E: the marble is neither green nor blue (A) ∴ P(E) = P(A) = 1/5 Let event F: the marble is green or red (not blue) ∴ P(F) = P (not C) = 1 – P(C) = 8/15

EXAMPLE 3

Number of sample points = n (S) = 36

Solution  A:

Two 3’s = {(3, 3)} n (A) = 1 and P(A) = 1/36  B: A 5 and a 6 = {(5, 6), (6, 5)} n (B) = 2 and P(B) = 1/18  C: A 5 or a 6 = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)} n (C) = 18 and P(C) = 1/2

Solution.  D:

At least one 6 = {(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} n (D) = 11 and P(D) = 11/36  E: Exactly one 6 = {(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)} n (E) = 10 and P(E) = 5/18  F: No 6’s = not D P(F) = 1 – P(D) = 1 – 11/36 = 25/36

Solution..  G:

Sum of 7 = {(1, 6), (2, 5), (3, 4), (4, 3),(5, 2), (6, 1)} n (G) = 6 and P(G) = 1/6  H: Sum is greater than 8 = sum is either 9 or 10 or 11. = {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)} n (H) = 10 and P(H) = 5/18

Solution…  I:

Sum of 7 or 11 = {(1, 6), (2, 5), (3, 4), (4, 3),(5, 2), (6, 1), (5, 6), (6, 5)} n (I) = 8 and P(I) = 2/9  J: Sum is no more than 8 = Not H P(J) = 1 - P(H) = 1 - 5/18 = 13/18

EXAMPLE 4 One lottery ticket is drawn at random from a set of 20 tickets numbered 1, 2, …, 20. Find the probability that the number on the ticket drawn is divisible by 3 or 5. Here n (S) = 20. Let A: number is divisible is divisible by 3 or 5 A = {3, 5, 6, 9, 10, 12, 15, 18, 20} n (A) = 9 P(A) = 9/20

 If

the letters of the word THURSDAY be arranged at random, what is the probability that the arrangement (i) begins with T (ii) begins with T and ends with U.

 Two

cards are drawn from a well shuffled pack of cards. what is the probability that (i) both are red cards, (ii) both are picture cards , (iii) one is a heart card and the other is club, (iv) both are kings, (v) one of them is an ace card.

 In

a batch of 400 bolts, 50 bolts are found to be defective. A bolt is selected from the batch. Find the probability that it is nondefective.  A bag contains 3 red, 4 blue and 5 green balls. Three balls are drawn at random. Find the probability that (i) all are blue, (ii) all are of the same colour, (iii) no ball is green, (iv) one ball is red, (v) two balls are blue.

 In

a class of 100 students, 70 failed in Maths, 55 failed in Physics and 22 failed in both. One student is selected at random. Find the probability that he fails (i) in both subjects, (ii) at least in one subject, (iii) neither of the subjects.  A room has 3 sockets. From a collection of a dozen bulbs of which 4 are defective, 5 are selected at random and put in sockets. Find the probability that the room (i) is dark, (ii) is lighted, (iii) lighted to its maximum.

FORMULAE  Events

A and B are independent if the happening of A does not affect the happening or non-happening of B and P(A and B) = P(A) × P(B)  Events A and B are dependent if the happening of A does affect the happening or non-happening of B and P(AB) = P(A and B) ÷ P(B)

EXAMPLE 1  Two

students A and B try to solve a problem. Probability of A solving is 3/5 and of B, 1/3. If they try independently what is the probability that the problem is not solved?  Given P(A) = 3/5 and P(B) = 1/3 P(A and B) = P(A) × P(B) = 3/5 × 1/3 = 1/5 P(A or B) = P(A) + P(B) – P(A and B) = 3/5 + 1/3 – 1/5 = 11/15 Thus P (neither A nor B) = 1 – 11/15 = 4/15

EXAMPLE 2 A

purse contains 4 silver and 5 gold coins. Another purse contains 6 silver and 4 gold coins A purse is selected and a coin is drawn. What is the probability that the coin is silver?  P (selecting each purse) = P(A) = P(B) = 1/2 P (coin is silver) = P (silver coin is drawn from A or B) = P(A) × P(SA) + P(B) × P(SB) = ½ × 4/9 + ½ × 6/10 = 47/90

EXAMPLE 3 A

purchasing agent has placed rush orders for a particular raw material with two different suppliers A and B. If neither order arrives in 4 days, the production process must shut down until at least one of the orders arrives. The probability that A delivers the material in 4 days is 0.55 and that B is 0.35. What is the probability that: (i) both the suppliers deliver material in 4 days? (ii) at least one supplier delivers the material in 4 days? (iii) the production will be shut?

Solution  Given:

P(A) = P (A supplies in 4 days) = 0.55 P(B) = P (B supplies in 4 days) = 0.35 Thus, P (not A) = 0.45 and P (not B) = 0.65 (i) P (both deliver material in 4 days) = 0.55 × 0.35 = 0.1925 (ii) P (at least one delivers the material in 4 days) = P (A but not B) + P (not A but B) = 0.55 × 0.65 + 0.45 × 0.35 = 0.515

Solution (iii) P (the production will be shut) = P (not A and not B) = 0.45 × 0.65 = 0.2925 There is 29.25% chance that the production will be shut because of unavailability of the raw material.

EXAMPLE 4  Small

cars get better mileage but are not as safe as big cars. Small cars accounted for 18% of the vehicles on the road but involved in 11898 deaths in accidents. The probability of an accident involving a small car leading to fatalities is 0.128 and the probability of an accident not involving a small car leading to fatalities is 0.05. Suppose you learn an accident involving a fatality, what is the probability a small car was involved?

Given: P(S) = P (small car on road) = 0.18 P (not S) = 1 – 0.18 = 0.82 P (fatal when small car) = P (FS) = 0.128 P (fatal when not small car) = P (Fnot S) = 0.05.

P (fatal accident with small car) = P (S and F) = 0.18 × 0.128 = 0.02304 P (fatal accident) P(F) = P (S and F) + P (not S and F) = 0.18 × 0.128 + 0.82 × 0.05 = 0.06404 P (small car involved  fatal accident) = P (SF) = P (S and F) / P(F) = 0.02304/0.06404 = 0.36

EXAMPLE 5 A

local bank is reviewing its credit card policy with a view toward recalling some of its credit cards. In the past approximately 5% of the card holders have defaulted and the bank has been unable to collect the outstanding balance. Hence, management has established a prior probability of 0.05 that any particular card holder will default. The bank has further found that the probability of missing one or more payments is 0.2 for customers who do not default. Of course the probability of missing one or more payments for those who default is 1.

EXAMPLE 5..  Given

that the customer has missed a payment, compute the probability that the customer will default.  The bank would like to recall its card if the probability that a customer will default is greater than 0.2. Should the bank recall its card if the customer misses a monthly payment? Why or why not?

Tree diagram

Given: P(D) = P (customer defaults) = 0.05 P (not D) = 1 – 0.05 = 0.95 Also, P (misses  defaults) = P (MD) = 1 and P (misses  not default) = P (Mnot D) = 0.2 P (D and M) = P(D) × P (M D) = 0.05 × 1 = 0.05 P(M) = P(D) × P (M D) + P (not D) × P (M not D) = 0.05 × 1 + 0.95 × 0.2 = 0.24 P (DM) = 0.05 / 0.24 = 0.21

EXERCISE

EXERCISE

A

game is played using a regular 12-faced fair die, with faces labelled 1 to 12, a coin and a simple board with nine squares as shown in the diagram. Initially the coin is placed on the shaded rectangle, die is rolled and if the outcome is is prime then coin is moved one place to R; otherwise it is towards L. The game stops when the coin reaches either R or L. Find the probability that the game (i) ends on 4th move at R, (ii) ends on 4th move, (iii) ends on 5th move, (iv) takes more than six moves.

 In

a certain part of the world there are more wet days than dry days. If a given day is wet, the probability that the next day will also be wet is 0.8. If a given day is dry, the probability that the following day will also be dry is 0.6. Give that Wednesday of a week is dry, calculate the probability that (i) Thursday, Friday of the same week are both wet, (ii) Friday of that week is wet, (iii) In a season, there were 44 matches played over 3 consecutive days with first and third days were dry. How many of these matches expect second day as wet?

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