Chapter 16 – Probability

Class 11 Mathematics · 44 questions · 0 with answers

Solved examples

example-1Multiple choice

An ordinary deck of cards contains 52 cards divided into four suits. The red suits are diamonds and hearts and black suits are clubs and spades. The cards J, Q, and K are called face cards. Suppose we pick one card from the deck at random.

(a)What is the sample space of the experiment?
(b)What is the event that the chosen card is a black face card? Solution (a) The outcomes in the sample space S are 52 cards in the deck. (b) Let E be the event that a black face card is chosen. The outcomes in E are Jack, Queen, King or spades or clubs. Symbolically E = {J, Q, K, of spades and clubs} or E = {J♣, Q♣, K♣, J♠, Q♠, K♠}

example-2Multiple choice

Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children. (a) List the eight elements in the sample space whose outcomes are all possible genders of the three children. (b) Write each of the following events as a set and find its probability :

(i)The event that exactly one child is a girl.
(ii)The event that at least two children are girls
(iii)The event that no child is a girl Solution (a) All possible genders are expressed as : S = {BBB, BBG, BGB, BGG, GBB,GBG, GGB, GGG} (b) (i)Let A denote the event : ‘exactly one child is a girl’ A = {BBG, BGB, GBB} P (A) = (ii) Let B denote the event that at least two children are girls. B = {GGB, GBG, BGG, GGG}, P (B) = . (iii) Let C denote the event : ‘no child is a girl’. C = {BBB} ∴ P (C) =

example-3Multiple choice

(a)How many two-digit positive integers are multiples of 3?
(b)What is the probability that a randomly chosen two-digit positive integer is a multiple of 3? Solution (a) 2 digit positive integers which are multiples of 3 are 12, 15, 18, ... , 99. Thus, there are 30 such integers. (b) 2-digit positive integers are 10, 11, 12, ... , 99. Thus, there are 90 such numbers. Since out of these, 30 numbers are multiple of 3, therefore, the probability that a 30 1 randomly chosen positive 2-digit integer is a multiple of 3, is = . 90 3

example-4Long answer

A typical PIN (personal identification number) is a sequence of any four symbols chosen from the 26 letters in the alphabet and the ten digits. If all PINs are equally likely, what is the probability that a randomly chosen PIN contains a repeated symbol? Solution A PIN is a sequence of four symbols selected from 36 (26 letters + 10 digits) symbols. By the fundamental principle of counting, there are 36 × 36 × 36 × 36 = 364 = 1,679,616 PINs in all. When repetition is not allowed the multiplication rule can be applied to conclude that there are

example-5Multiple choice

An experiment has four possible outcomes A, B, C and D, that are mutually exclusive. Explain why the following assignments of probabilities are not permissible:

(a)P(A) = .12, P
(B)= .63, P
(C)= 0.45, P
(D)= – 0.20 9 45 27 46 (b) P(A) = , P (B) = P (C) = P (D) = 120 120 120 120 Solution (a) Since P(D) = – 0.20, this is not possible as 0 ≤ P (A) ≤ 1 for any event A. 9 45 27 46 127 (b) P(S) = P (A ∪ B ∪ C ∪ D) = + + + = ≠1 . 120 120 120 120 120 This violates the condition that P (S) = 1.

example-6Long answer

Probability that a truck stopped at a roadblock will have faulty brakes or badly worn tires are 0.23 and 0.24, respectively. Also, the probability is 0.38 that a truck stopped at the roadblock will have faulty brakes and/or badly working tires. What is the probability that a truck stopped at this roadblock will have faulty breaks as well as badly worn tires? Solution Let B be the event that a truck stopped at the roadblock will have faulty brakes and T be the event that it will have badly worn tires. We have P (B) = 0.23, P (T) = 0.24 and P (B∪T) = 0.38 and P (B ∪ T) = P (B) + P (T) – P (B ∩ T) So 0.38 = 0.23 + 0.24 – P (B ∩ T) ⇒ P (B ∩ T) = 0.23 + 0.24 – 0.38 = 0.09

example-7Long answer

If a person visits his dentist, suppose the probability that he will have his teeth cleaned is 0.48, the probability that he will have a cavity filled is 0.25, the probability that he will have a tooth extracted is 0.20, the probability that he will have a teeth cleaned and a cavity filled is 0.09, the probability that he will have his teeth cleaned and a tooth extracted is 0.12, the probability that he will have a cavity filled and a tooth extracted is 0.07, and the probability that he will have his teeth cleaned, a cavity filled, and a tooth extracted is 0.03. What is the probability that a person visiting his dentist will have atleast one of these things done to him? Solution Let C be the event that the person will have his teeth cleaned and F and E be the event of getting cavity filled or tooth extracted, respectively. We are given P(C) = 0.48, P (F) = 0.25, P (E) = .20, P (C ∩ F) = .09, P (C ∩ E) = 0.12, P (E ∩ F) = 0.07 and P (C ∩ F ∩ Ε) = 0.03 Now, P ( C ∪ F ∪ E) = P (C) + P (F) + P (E) – P (C ∩ F) – P (C ∩ E) – P (F ∩ E) + P (C ∩ F ∩ E) = 0.48 + 0.25 + 0.20 – 0.09 – 0.12 – 0.07 + 0.03 = 0.68

example-8Multiple choice

An urn contains twenty white slips of paper numbered from 1 through 20, ten red slips of paper numbered from 1 through 10, forty yellow slips of paper numbered from 1 through 40, and ten blue slips of paper numbered from 1 through 10. If these 80 slips of paper are thoroughly shuffled so that each slip has the same probability of being drawn. Find the probabilities of drawing a slip of paper that is

(a)blue or white
(b)numbered 1, 2, 3, 4 or 5
(c)red or yellow and numbered 1, 2, 3 or 4
(d)numbered 5, 15, 25, or 35; (e) white and numbered higher than 12 or yellow and numbered higher than 26. Solution (a) P (Blue or White) = P (Blue) + P (White) (Why?) 10 20 30 3 = + = = 80 80 80 8 (b) P (numbered 1, 2, 3, 4 or 5) = P (1 of any colour) + P (2 of any colour) + P (3 of any colour) + P (4 of any colour) + P (5 of any colour) 4 4 4 4 4 20 2 1 = + + + + = = = 80 80 80 80 80 80 8 4 (c) P (Red or yellow and numbered 1, 2, 3 or 4) = P (Red numbered 1, 2, 3 or 4) + P (yellow numbered 1, 2, 3 or 4) PROBABILITY 293 4 4 8 1 = + = = 80 80 80 10 (d) P (numbered 5, 15, 25 or 35) = P (5) + P (15) + P (25) + P (35) = P (5 of White, Red, Yellow, Blue) + P (15 of White, Yellow) + P (25 of Yellow) + P (35 of Yellow) 4 2 1 1 8 1 = + + + = = 80 80 80 80 80 10 (e) P (White and numbered higher than 12 or Yellow and numbered higher than 26) = P (White and numbered higher than 12) + P (Yellow and numbered higher than 26) 8 14 22 11 = + = = 80 80 80 40

example-9Multiple choice

In a leap year the probability of having 53 Sundays or 53 Mondays is 2 3 4 5

(A)
(B)
(C)
(D)7 7 7 7 Solution (B) is the correct answer. Since a leap year has 366 days and hence 52 weeks and 2 days. The 2 days can be SM, MT, TW, WTh, ThF, FSt, StS. Therefore, P (53 Sundays or 53 Mondays) = .

example-10Multiple choice

Three digit numbers are formed using the digits 0, 2, 4, 6, 8. A number is chosen at random out of these numbers. What is the probability that this number has the same digits? 1 16 1 1

(A)
(B)
(C)
(D)16 25 645 25 Solution (D) is the correct answer. Since a 3-digit number cannot start with digit 0, the hundredth place can have any of the 4 digits. Now, the tens and units place can have all the 5 digits. Therefore, the total possible 3-digit numbers are 4 × 5 × 5, i.e., 100. The total possible 3 digit numbers having all digits same = 4 4 1 Hence, P (3-digit number with same digits) = = . 100 25

example-11Multiple choice

Three squares of chess board are selected at random. The probability of getting 2 squares of one colour and other of a different colour is 16 8 3 3

(A)
(B)
(C)
(D)21 21 32 8 Solution (A) is the correct answer. In a chess board, there are 64 squares of which 32 are white and 32 are black. Since 2 of one colour and 1 of other can be 2W, 1B, or 1W, 2B, the number of ways is (32C2 × 32C1) × 2 and also, the number of ways of choosing any 3 boxes is 64C3. C 2 × 32 C1 × 2 16 Hence, the required probability = 64 = . C3 21 1 2

example-12Multiple choice

If A and B are any two events having P (A ∪ B) = and P ( A ) = , 2 3 then the probability of A ∩ B is 1 2 1 1

(A)
(B)
(C)
(D)2 3 6 3 Solution (C) is the correct answer. We have P(A ∪ B) = ⇒ P (A ∪ (B – A)) = ⇒ P (A) + P (B – A) = (since A and B – A are mutually exclusive) ⇒ 1 – P ( A ) + P (B – A) = 2 1 ⇒ 1– + P (B – A) = 3 2 PROBABILITY 295 ⇒ P (B – A) = ⇒ P ( A ∩ B) = (since A ∩ B ≡ B – A)

example-13Multiple choice

Three of the six vertices of a regular hexagon are chosen at random. What is the probability that the triangle with these vertices is equilateral? 3 3 1 1

(A)
(B)
(C)
(D)10 20 20 10 Solution (D) is the correct answer. F B E C Fig. 16.1 ABCDEF is a regular hexagon. Total number of triangles 6C3 = 20. (Since no three points are collinear). Of these only ∆ ACE; ∆ BDF are equilateral triangles. 2 1 Therefore, required probability = = . 20 10

example-14Multiple choice

If A, B, C are three mutually exclusive and exhaustive events of an experiment such that 3P(A) = 2P(B) = P(C), then P(A) is equal to 1 2 5 6

(A)
(B)
(C)
(D)11 11 11 11 Solution (B) is the correct answer. Let 3P (A) = 2P(B) = P(C) = p which gives p (A) p p = , P(B) = and P(C) = p 3 2 Now since A, B, C are mutually exclusive and exhaustive events, we have P(A) + P(B) + P(C) = 1 p p 6 ⇒ + + p =1 ⇒ p= 3 2 11 p 2 Hence, P (A) = = 3 11

example-15Multiple choice

One mapping (function) is selected at random from all the mappings of the set A = {1, 2, 3, ..., n} into itself. The probability that the mapping selected is one to one is 1 1 n −1

(A)
(B)
(C)
(D)none of these nn n n n −1 Solution (C) is the correct answer. Total number of mappings from a set A having n elements onto itself is nn Now, for one to one mapping the first element in A can have any of the n images in A; the 2nd element in A can have any of the remaining (n – 1) images, counting like this, the nth element in A can have only 1 image. Therefore, the total number of one to one mappings is n . n n n −1 n −1 Hence the required probability is = n −1 = . nn n n −1

Questions

Q36Short answer

× 35 × 34 × 33 = 1,413,720 different PINs PROBABILITY 291 The number of PINs that contain at least one repeated symbol = 1,679,616 – 1,413,720 = 2,65,896 Thus, the probability that a randomly chosen PIN contains a repeated symbol is 265,896 = .1583 1, 679, 616

Q1Short answer

If the letters of the word ALGORITHM are arranged at random in a row what is the probability the letters GOR must remain together as a unit?

Q2Short answer

Six new employees, two of whom are married to each other, are to be assigned six desks that are lined up in a row. If the assignment of employees to desks is made randomly, what is the probability that the married couple will have nonadjacent desks? [Hint: First find the probability that the couple has adjacent desks, and then subtract it from 1.]

Q3Short answer

Suppose an integer from 1 through 1000 is chosen at random, find the probability that the integer is a multiple of 2 or a multiple of 9.

Q4Multiple choice

An experiment consists of rolling a die until a 2 appears.

(i)How many elements of the sample space correspond to the event that the 2 appears on the kth roll of the die? PROBABILITY 297
(ii)How many elements of the sample space correspond to the event that the 2 appears not later than the kth roll of the die? [Hint:(a) First (k – 1) rolls have 5 outcomes each and kth rolls should result in 1 outcomes. (b)1 + 5 + 52 + ... + 5k–1.]

Q5Short answer

A die is loaded in such a way that each odd number is twice as likely to occur as each even number. Find P(G), where G is the event that a number greater than 3 occurs on a single roll of the die.

Q6Short answer

In a large metropolitan area, the probabilities are .87, .36, .30 that a family (randomly chosen for a sample survey) owns a colour television set, a black and white television set, or both kinds of sets. What is the probability that a family owns either anyone or both kinds of sets?

Q7Multiple choice

If A and B are mutually exclusive events, P

(A)= 0.35 and P
(B)= 0.45, find (a) P (A′) (b) P (B′)
(c)P (A ∪ B)
(d)P (A ∩ B) (e) P (A ∩ B′) (f) P (A′ ∩ B′)

Q8Multiple choice

A team of medical students doing their internship have to assist during surgeries at a city hospital. The probabilities of surgeries rated as very complex, complex, routine, simple or very simple are respectively, 0.15, 0.20, 0.31, 0.26, .08. Find the probabilities that a particular surgery will be rated

(a)complex or very complex;
(b)neither very complex nor very simple;
(c)routine or complex
(d)routine or simple

Q9Multiple choice

Four candidates A, B, C, D have applied for the assignment to coach a school cricket team. If A is twice as likely to be selected as B, and B and C are given about the same chance of being selected, while C is twice as likely to be selected as D, what are the probabilities that

(a)C will be selected?
(b)A will not be selected?

Q10Multiple choice

One of the four persons John, Rita, Aslam or Gurpreet will be promoted next month. Consequently the sample space consists of four elementary outcomes S = {John promoted, Rita promoted, Aslam promoted, Gurpreet promoted} You are told that the chances of John’s promotion is same as that of Gurpreet, Rita’s chances of promotion are twice as likely as Johns. Aslam’s chances are four times that of John.

(a)Determine P (John promoted) P (Rita promoted) P (Aslam promoted) P (Gurpreet promoted)
(b)If A = {John promoted or Gurpreet promoted}, find P (A).

Q11Multiple choice

The accompanying Venn diagram shows three events, A, B, and C, and also the probabilities of the various intersections (for instance, P (A ∩ B) = .07). Determine

(a)P (A)
(b)P (B ∩ C )
(c)P (A ∪ B)
(d)P (A ∩ B ) (e) P (B ∩ C) (f) Probability of exactly one of the three occurs.

Q12Multiple choice

One urn contains two black balls (labelled B1 and B2) and one white ball. A second urn contains one black ball and two white balls (labelled W1 and W2). Suppose the following experiment is performed. One of the two urns is chosen at random. Next a ball is randomly chosen from the urn. Then a second ball is chosen at random from the same urn without replacing the first ball.

(a)Write the sample space showing all possible outcomes
(b)What is the probability that two black balls are chosen?
(c)What is the probability that two balls of opposite colour are chosen?

Q13Multiple choice

A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the Probability that

(a)All the three balls are white
(b)All the three balls are red
(c)One ball is red and two balls are white

Q14Multiple choice

If the letters of the word ASSASSINATION are arranged at random. Find the Probability that

(a)Four S’s come consecutively in the word
(b)Two I’s and two N’s come together
(c)All A’s are not coming together
(d)No two A’s are coming together. PROBABILITY 299

Q15Long answer

A card is drawn from a deck of 52 cards. Find the probability of getting a king or a heart or a red card.

Q16Multiple choice

A sample space consists of 9 elementary outcomes e1, e2, ..., e9 whose probabilities P(e1) = P(e2) = .08, P(e3) = P(e4) = P(e5) = .1 P(e6) = P(e7) = .2, P(e8) = P(e9) = .07 SupposeA = {e1, e5, e8}, B = {e2, e5, e8, e9}

(a)Calculate P (A), P
(B), and P (A ∩ B) (b) Using the addition law of probability, calculate P (A ∪ B)
(c)List the composition of the event A ∪ B, and calculate P (A ∪ B) by adding the probabilities of the elementary outcomes.
(d)Calculate P ( B ) from P (B), also calculate P ( B ) directly from the elementary outcomes of B

Q17Multiple choice

Determine the probability p, for each of the following events.

(a)An odd number appears in a single toss of a fair die.
(b)At least one head appears in two tosses of a fair coin.
(c)A king, 9 of hearts, or 3 of spades appears in drawing a single card from a well shuffled ordinary deck of 52 cards.
(d)The sum of 6 appears in a single toss of a pair of fair dice.

Q18Multiple choice

In a non-leap year, the probability of having 53 tuesdays or 53 wednesdays is 1 2 3

(A)
(B)
(C)
(D)none of these 7 7 7

Q19Multiple choice

Three numbers are chosen from 1 to 20. Find the probability that they are not consecutive 186 187 188 18

(A)
(B)
(C)
(D)20 190 190 190 C3

Q20Multiple choice

While shuffling a pack of 52 playing cards, 2 are accidentally dropped. Find the probability that the missing cards to be of different colours 29 1 26 27

(A)
(B)
(C)
(D)52 2 51 51

Q21Multiple choice

Seven persons are to be seated in a row. The probability that two particular persons sit next to each other is 1 1 2 1

(A)
(B)
(C)
(D)3 6 7 2

Q22Multiple choice

Without repetition of the numbers, four digit numbers are formed with the numbers 0, 2, 3, 5. The probability of such a number divisible by 5 is 1 4 1 5

(A)
(B)
(C)
(D)5 5 30 9

Q23Multiple choice

If A and B are mutually exclusive events, then

(A)P (A) ≤ P ( B )
(B)P (A) ≥ P ( B )
(C)P (A) < P ( B )
(D)none of these

Q24Multiple choice

If P (A ∪ B) = P (A ∩ B) for any two events A and B, then

(A)P (A) = P
(B)(B) P (A) > P (B)
(C)P (A) < P (B)
(D)none of these

Q25Multiple choice

6 boys and 6 girls sit in a row at random. The probability that all the girls sit together is 1 12 1

(A)
(B)
(C)
(D)none of these 432 431 132

Q26Multiple choice

A single letter is selected at random from the word ‘PROBABILITY’. The probability that it is a vowel is 1 4 2 3

(A)
(B)
(C)
(D)3 11 11 11

Q27Multiple choice

If the probabilities for A to fail in an examination is 0.2 and that for B is 0.3, then the probability that either A or B fails is

(A)> . 5
(B).5
(C)≤ .5
(D)0

Q28Multiple choice

The probability that at least one of the events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.2, then P ( A ) + P ( B ) is

(A)0.4
(B)0.8
(C)1.2
(D)1.6 PROBABILITY 301

Q29Multiple choice

If M and N are any two events, the probability that at least one of them occurs

(A)P (M) + P (N) – 2 P (M ∩ N)
(B)P (M) + P (N) – P (M ∩ N)
(C)P (M) + P (N) + P (M ∩ N)
(D)P (M) + P (N) + 2P (M ∩ N) State whether the statements are True or False in each of the Exercises 30 to 36.

Q30Multiple choice

The probability that a person visiting a zoo will see the giraffee is 0.72, the probability that he will see the bears is 0.84 and the probability that he will see both is 0.52.

Q31Multiple choice

The probability that a student will pass his examination is 0.73, the probability of the student getting a compartment is 0.13, and the probability that the student will either pass or get compartment is 0.96.

Q32Multiple choice

The probabilities that a typist will make 0, 1, 2, 3, 4, 5 or more mistakes in typing a report are, respectively, 0.12, 0.25, 0.36, 0.14, 0.08, 0.11.

Q33Multiple choice

If A and B are two candidates seeking admission in an engineering College. The probability that A is selected is .5 and the probability that both A and B are selected is at most .3. Is it possible that the probability of B getting selected is 0.7?

Q34Multiple choice

The probability of intersection of two events A and B is always less than or equal to those favourable to the event A.

Q35Multiple choice

The probability of an occurrence of event A is .7 and that of the occurrence of event B is .3 and the probability of occurrence of both is .4.

Q36Fill in the blanks

The sum of probabilities of two students getting distinction in their final examinations is 1.2. Fill in the blanks in the Exercises 37 to 41.

Q37Fill in the blanks

The probability that the home team will win an upcoming football game is 0.77, the probability that it will tie the game is 0.08, and the probability that it will lose the game is _____.

Q38Fill in the blanks

If e1, e2, e3, e4 are the four elementary outcomes in a sample space and P(e1) = .1, P(e2) = .5, P (e3) = .1, then the probability of e4 is ______.

Q39Fill in the blanks

Let S = {1, 2, 3, 4, 5, 6} and E = {1, 3, 5}, then E is _________.

Q40Fill in the blanks

If A and B are two events associated with a random experiment such that P ( A) = 0.3, P (B) = 0.2 and P (A ∩ B) = 0.1, then the value of P (A ∩ B ) is _______.

Q41Fill in the blanks

The probability of happening of an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then the probability of neither A nor B is ________.

Q42Multiple choice

Match the proposed probability under Column C1 with the appropriate written description under column C2 : C1 C2 Probability Written Description (a) 0.95

(i)An incorrect assignment (b) 0.02
(ii)No chance of happening (c) – 0.3
(iii)As much chance of happening as not. (d) 0.5
(iv)Very likely to happen (e) 0 (v) Very little chance of happening

Q43Multiple choice

Match the following (a) If E1 and E2 are the two mutually

(i)E1 ∩ E2 = E1 exclusive events (b) If E1 and E2 are the mutually
(ii)(E1 – E2) ∪ (E1 ∩ E2) = E1 exclusive and exhaustive events (c) If E1 and E2 have common
(iii)E1 ∩ E2 = φ, E1 ∪ E2 = S outcomes, then (d) If E1 and E2 are two events
(iv)E1 ∩ E2 = φ such that E1 ⊂ E2