## Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 16 Probability

**Short Answer** **Type Questions **

**(Score 3)**

Question 1.

A basket contains three apples. What is the sample space if,

a. One apple is chosen

b. Two apples are chosen

c. Three apples are chosen.

Answer:

Let the Apples be A, B and C

- S
_{1}= {A,B,C} - S
_{2}={AB,AC,BC} - S
_{3}={ABC}

Question 2.

a. A coin is tossed again and again till head turns up. Write down the sample space.

b. From each of the four married couples, one of the partners is selected at random. Find the probability that those selected are of the same sex.

c. what is the probability of drawing a‘king’ from a well-shuffled deck of 52 cards ?

Answer:

a. S = {H, TH, TTH, TTTH, ………………….}

b. Required probability =

c. Well-shuffled ensures equally likely outcomes. There are 4 kings in a deck. Thus

P(a king) =

Question 3.

If three fair coins are tossed,

i. Write down the sample space.

ii. What is the probability of getting exactly two heads?

iii. Estimate the probability of getting at least one head.

Answer:

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

ii. Favourable cases are HHT, HTH, THH

∴ Required probability =

iii. P (At least one head) = 1 – P (no head)

=

Question 4.

A card is drawn at random from a well shuffled pack of 52 cards. Find the probability of its being a spade or a king.

Answer:

P (spade or king) = P (spade) + P (king) – P (king of spade)

=

Question 5.

A die has two faces each with number 1, three faces with number 2 and one face with number 3. If the die is rolled once, determine,

(i) P (2)

(ii) P (1 or 3)

(iii) P (not 3)

Answer:

There are 6 outcomes.

i. A: getting 2 can be obtained in 3 ways

ii. B: getting 1 or 3. This can be obtained in 3ways

iii. C: not 3. This can be obtained in 5 ways P (C) = 5/6

Question 6.

If A and B are two mutually exclusive and exhaustive events, then

i. P (A) + P (B) =…………….. [0,1,1/2]

ii. A fair die is thrown. What is the probability that either an odd number or a number greater than 4 will turn up.

Answer:

i.P(A) + P(B)= 1 .

ii Let A – Event that an odd number turn up B – Event that a number greater than 4 will turn up.

P (A∪B) = P (A) + P (B) – P (A ∩B)

Question 7.

The probabilities that a student will receive A, B, C and D grade are 0.4,0.35,0.15 and

0,1 respectively. Find the probability that a student will receive

i. not an A grade

ii. B or C grade

iii. Atmost C grade

Answer:

(A)= 0.4, P (B)=0.35, P (C)=0.15,P (D)=0.1

i. P (not A grade) = P (A’) = 1 – P (A)

= 1-0.4 = 0.6

ii. P (B or C grade) = P(B) + P(C)

= 0.35 + 0.15 = 0.5

iii. P (At most C grade) = P (C or D)

= 0.15 + 0.1=0.25

Question 8.

A card is drawn from a well shuffled deck of 52 cards. Find the probability of getting (i) a heart (ii) a seven of heart.

Answer:

i. There are 13 cards of heart in a deck of 52 cards

P[getting a heart] =

ii. There is only one seven of heart in a deck.

P[ getting a seven of heart]=

Question 9.

A pair disc is thrown. Let A be the event “ the sum is greater than 8” and B be the event “ 2 occurs on either die”. Are A and B mutually exclusive ?

Answer:

Here the sample space

S ={(1,1), (1,2),……. (6,6)}

A = the sum is greater than 8

= {(3,6),(4,5), (5,4) ,(6,3), (4,6), (6,4), (5,5), (6,5), (5,6), (6,6)}

B = 2 occurs on either die

= {(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (1,2),(4,2), (5,2), (6,2)}

A∩B=φ

A and B are mutually exclusive.

**Short Answer** **Type Questions **

**(Score 4)**

Question 1.

a. Probability of having atleast one tail in 4 throws with a coin is

a.

b.

c.

d. 1

b. In a non-leap year, find the probability of getting 53 Sundays or 53 Tuesdays or 53 Thursday.

Answer:

a.

There are 2^{4} outcomes on 4 tosses of a coin

2^{4}= 16 outcomes

Probability of having atleast one tail = 1-probability of getting no. of tails on 4

tosses of a coin =

(only 1 HHH contains no tails. The chances of no. tails are

b. A non leap year has 365 days. There are 52 weeks each having 7 days. 52 weeks corresponds to 7 x 52 = 364 days The remaining 1 day can be day among monday, tuesday,…. Sunday. Hence required probability.

Question 2.

Find the probability that,

a. non leap year should have 53

b. leap year should have 53

Answer:

a. In an ordinary year there are 52 complete weeks and one extra day. i.e, 365 = 52 x 7 + 1. If that extra day is Sunday, there will be 53 Sundays. Since there are 7 equally likely days.

Required probability = 1/7

b. A leap year has 52 complete weeks and 2 extra days. Extra days can be [(sun, mon), (mon, tue), (tue, wed), (wed, thu), (thu, fri), (fri, sat), (sat, sun)] of which 2 are favourable cases. Therefore required robability = 2/7

Question 3.

In a lottery of 50 tickets number from 1 to 50, two tickets are drawn simultaneously.Find the probability that-

i. Both the tickets drawn have prime numbers.

ii. None of the tickets drawn has prime number.

Answer:

i. Prime numbers between 1 and 50 are 2, 3, 5,7,11,13,17,19,23,29,31,37,41,43,47

i..e, 15 prime numbers

Required probability

Question 4.

A problem is given to three students whose chances of solving it are 1/2, 1/3 and 1/4 respectively. What is the probability that

i. None solve the problem.

ii. At least one solve the problem.

Answer:

Question 5.

a. A pair of dice is thrown. Find the probability of getting a doublet

b. Three letters are written to different persons and addresses on the envelope are also written. Without looking at the addresses, the letters are put into the envelopes, the probability that letters go into right envelopes.

a.

b.

c.

d.

Answer:

Here 36 sample points are there.

Let A: getting a doublet A= {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

P (Doublet) =

b.

Total number of ways in which letters can be put in envelopes is 3×2×1= 6 ways If two are sent to correct addresses, it implies the third one will also be sent to correct address. This gives 1 such way.

So, the required probability =

Question 6.

a. If P(A) = 0.5, P(B) = 0.2, then find P(A∪B)’. (Given that A and B are mutually exclusive).

b A pair of dice is thrown, then estimate the probability of ‘ getting an even number on the first die’.

Answer:

Question 7.

A bag contains 9 balls of which 4 are red, 3 are white and 2 are green. If a ball is drawn at random, then:

i. Calculate the probability that it is not white.

ii. Calculate the probability that it is either white or green.

Answer:

i. P(getting a white ball)

∴ P(not a white ball) = 1 – P(getting a white ball) =

ii. P(either a white or green)

**Long Answer** **Type Questions **

**(Score 6)**

Question 1.

A problem in mathematics is given to 3 students whose chances of solving individually are .Find the probability that the problem will be solved atleast by one.

Answer:

Let A, B, C be the respective events of solving the problem and be the respective events of not solving the problem. Then A, B, C are independent events.

Question 2.

Let A and B be two events, not mutually exclusive, connected with a random experiment

If , and . Find the values of the following probabilities.

i. P(A∩B)

ii.P(A∩B^{c})

iii.P(A∩B)^{c }Answer:

ii. The event A will occur if and only if any one of two mutually exclusive events

Question 3.

A husband and wife appear for an interview for two vacancies in the same post. The probability of husbands selection is 1/7 and that of wife’s selection is 1/5. What is the probability that

i. Both of them will be selected.

ii. Only one of them will be selected.

iii. none of them will be selected.

iv. At least one of them will be selected.

Answer:

Question 4.

A speaks truth in 75% and B in 80% of the cases. In what percentage of cases are they likely to contradict each other in narrating the same incident.

Answer:

Question 5.

If the letter of the word ‘ATTRACTION’ are written down at random, find the probability that

i. All the T’s occur together

ii. No two T’s occur together

Answer:

Total number of arrangements of the letters of the word ATTRACTION is

i. Considering three T’s as one letter. There are 8 letters which can be arranged in ways.

Required probability

ii. Other than 3 T’s there are 7 letters which 7! can be arranged in

There are 8 places, 6 between the 7 letters and one each on extreme left and right. To separate three T’s, we arrange them in these 8 places which can be done in 8C_{3} ways.

Number of ways in which no two T’s

Question 6.

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

i. All the three balls are white.

ii. All the three balls are red.

iii. One ball is red and two balls are white

b. There are 19 balls numbered from 1 to 19 in a bag. If a person selects one at random what is the probability that the number printed on the ball be an even number greater than 9 ?

Answer:

b. Total number = 19

The numbers 9 as well as even are 10,12, 14,16,18.

Favourable cases = 5 P [ Number is even and greater than 9] =

Question 7.

a. If A and B are two events in a random experiment, then P(A) + P(B) – P(A∩B)

b. Given P(A) = 0.5, P(B) = 0.6 and P(A∩B) =03. Find P(A ∪ B) and P(A’). c. Two dice are thrown simultaneously. Find the probability of getting a doublet (same number on both dice).

Answer:

Question 8.

a. If P (A) = 0.5, P (B) = 0.2, find P(A ∪ B)’ (Given that A and B are mutually exclusive).

b. A pair of dice is thrown. Then estimate the event of getting an even number on the first die.

c. A bag contains 9 balls of which 4 are red, 3 are white and 2 are green. If a ball is drawn at random then:

i. Calculate the probability that it is not white.

ii. It is either white or green.

Answer:

a.P (A∪B) = P (A) + P (B) = 0.5 + 0.2 = 0.7

P(A∪B)’ = l-P(A∪B)= 1-0.7 = 03

b. Events are (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (6,1)(6,2) (6,4) (6,5) (6,6)

Question 9.

i. Three coins are tossed once. Write the sample space of the experiment.

ii. One card is drawn from a well shuffled pack of 52 cards. If each outcome is equally likely, calculate the probability that the card will be a diamond.

iii. If A and B are two events of a random experiment such that P(A)= 1/4, P (B) =1/ 2 and

P (A and B) = 1/8, then find P (A or B) and P (not A and not B).

Answer:

i.

ii Required probability =

iii. P (A or B) = P (A) + P (B) – P (A∩B)=

P (not A and not B) = P (A’∩B’)

= 1 -P(A∪B) = 3/8

**NCERT Questions and Answers**

Question 1.

One card is drawn at random from a pack of 52 cards. Find the probability that

a. The card drawn is red.

b. The card drawn is an ace.

c. The card drawn is a red ace.

d The card drawn is red or an ace.

e. The card drawn is neither red nor ace.

Answer:

Question 2.

Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that

a. Both Anil and Ashima will not qualify the examination.

b. Atleast one of them will not qualify the examination.

c. Only one of them will qualify the examination.

Answer:

Let E and F denote the events that Anil and Ashima will qualify the examination, respectively. Given that

P (E) = 0.05, P (F) = 0.10 and P (E ∩F) = 0.02.

a. The event ‘both Anil and Ashima will not qualify the examination’ may be expressed as E’∩F’.

Since, E’ is ‘not E’, i.e., Anil will not qualify the examination and F’ is ‘not F’, i.e.,

Ashima will not qualify the examination.

Also E’ ∩F’=(E ∩F)’ (by Demorgan’s law)

Question 3.

In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both. If one of these students is selected at random, find the probability that

i. The student opted for NCC or NSS.

ii. The student has opted neither NCC nor NSS.

iii. The student has opted NSS but not NCC. Let the events be A and B.

Answer:

Let the events be A and B

Question 4.

A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that

i. All will be blue.

ii. At least one will be green.

Answer:

Question 5.

A die is drawn thrown, find the probability of following events.

i. A prime number will appear.

ii. A number greater than or equal to 3 will appear.

iii. A number less than or equal to one will appear.

iv. A number more than 6 will appear.

v. A number less than 6 will appear.

Answer:

i. When a die is thrown, the sample space associated to the random experiment = {1, 2,3,4,5,6} Prime numbers are 2,3,5

Therefore, probability of getting 2,3 or 5

ii. Numbers grater than or equal to 3 are 3,4,5,6

Therefore, probability of getting 3,4,5 or 6

iii. A number less or equal to one is only 1. Therefore probability of getting

iv. Probability of more than 6 will appear =P(φ)=o

Number less than 6 are 1, 2, 3, 4 or 5. Therefore, probability of getting 1,2,3,4 or 5 are

Question 6.

A card is selected from a pack of 52 cards.

a. How many points are there in the sample space?

b. Calculate the probability that the card is an ace of spades.

c. Calculate the probability that the card is

(i) an ace

(ii) black card.

Answer:

a. Since a card is selected from a pack of 52 cards, the number of points in the sample space is 52.

b. There is one ace in spade. Total cards = 52

∴ Probability of one ace in spade =

c. i.There are 4 ace in 52 cards,

∴ Probability of one ace =

ii. There are 26 black cards in a pack of 52 cards.

∴ Probability of getting one black card

Question 7.

A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed, find the probability that the sum of numbers that turn up is (i) 3 (ii) 12.

Answer:

The number of outcomes are (1,1),(1,2),(1,3),(1,5), (1,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6).

i. The favourable outcomes are (1,2)

Number of total outcomes 12 and number of favourable outcomes 1.

∴ Probability of getting sum 3 =

ii. The favourable outcomes are (6,6).

∴ Probability of getting sum 12=

Question 8.

There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?

Answer:

‘There are 6 woman in total of 10 city council. Therefore probability of getting a woman council

Question 9.

A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is

i. a vowel

ii. a consonant.

Answer:

a. The elementary events are

A, A, I, A, I, O

∴ Required probability =

b. Favourable outcomes are S, S, S, S, N,

T, N ∴ Required probability =

#### Plus One Maths Chapter Wise Questions and Answers