Std 10 Mathematics Part 1 · Chapter 5

5. Probability

Solutions to Practice Sets 5.1 to 5.4 and Problem Set 5, with sample spaces, events and step-by-step probability working. Maharashtra State Board (Balbharati), Std 10 Mathematics Part 1 (Algebra).

Practice Set 5.1

1How many possibilities are there in each of the following?

(1) Vanita plans to visit one of these sites: Ajintha, Mahabaleshwar, Lonar Sarovar, Tadoba wild life sanctuary, Amboli, Raigad, Matheran, Anandavan.

Any one of the 8 sites can be selected.

8 possibilities

(2) Any day of a week is to be selected randomly.

Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday

7 possibilities

(3) Select one card from the pack of 52 cards.

Any one of the 52 cards can be selected.

52 possibilities

(4) One number from 10 to 20 is written on each card. Select one card randomly.

The numbers are 10, 11, 12, …, 20.

Count = 20 − 10 + 1 = 11

11 possibilities

Practice Set 5.2

1For each of the following experiments write sample space ‘S’ and number of sample points n(S).

(1) One coin and one die are thrown simultaneously.

S = {H1, H2, H3, H4, H5, H6,
T1, T2, T3, T4, T5, T6}

n(S) = 2 × 6 = 12

n(S) = 12

(2) Two digit numbers are formed using digits 2, 3 and 5 without repeating a digit.

S = {23, 25, 32, 35, 52, 53}

n(S) = 6

2The arrow is rotated and it stops randomly on the disc (colours Red, Orange, Yellow, Blue, Green, Purple). Find out on which colour it may stop.

S = {Red, Orange, Yellow, Blue, Green, Purple}

n(S) = 6

It may stop on any one of the 6 colours; n(S) = 6

3In the month of March 2019, find the days on which the date is a multiple of 5.

Date	5	10	15	20	25	30
Day	Tuesday	Sunday	Friday	Wednesday	Monday	Saturday

Dates that are multiples of 5: S = {5, 10, 15, 20, 25, 30}

n(S) = 6

5 March (Tue), 10 (Sun), 15 (Fri), 20 (Wed), 25 (Mon), 30 (Sat)

4Form a ‘Road safety committee’ of two, from 2 boys (B₁, B₂) and 2 girls (G₁, G₂). Complete the following activity to write the sample space.

(a) Committee of 2 boys = B₁B₂

(b) Committee of 2 girls = G₁G₂

∴ Sample space = {B₁B₂, G₁G₂, B₁G₁, B₁G₂, B₂G₁, B₂G₂}

n(S) = 6

Practice Set 5.3

1Write sample space ‘S’ and number of sample points n(S) for each of the following experiments. Also write events A, B, C in the set form and write n(A), n(B), n(C).

(1) One die is rolled. Event A: even number; Event B: odd number; Event C: prime number on the upper face.

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

A = {2, 4, 6}, n(A) = 3

B = {1, 3, 5}, n(B) = 3

C = {2, 3, 5}, n(C) = 3

(2) Two dice are rolled simultaneously. Event A: sum is a multiple of 6; Event B: sum is minimum 10; Event C: same digit on both upper faces.

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

n(S) = 36

A: sum is 6 or 12 → A = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6)}, n(A) = 6

B: sum is 10, 11 or 12 → B = {(4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}, n(B) = 6

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

(3) Three coins are tossed simultaneously. A: at least two heads; B: no head; C: head on the second coin.

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, n(S) = 8

A = {HHH, HHT, HTH, THH}, n(A) = 4

B = {TTT}, n(B) = 1

C = {HHH, HHT, THH, THT}, n(C) = 4

(4) Two digit numbers are formed using digits 0, 1, 2, 3, 4, 5 without repetition. A: number is even; B: divisible by 3; C: greater than 50.

The tens digit cannot be 0.

S = {10, 12, 13, 14, 15,
20, 21, 23, 24, 25,
30, 31, 32, 34, 35,
40, 41, 42, 43, 45,
50, 51, 52, 53, 54}

n(S) = 25

A = {10, 12, 14, 20, 24, 30, 32, 34, 40, 42, 50, 52, 54}, n(A) = 13

B = {12, 15, 21, 24, 30, 42, 45, 51, 54}, n(B) = 9

C = {51, 52, 53, 54}, n(C) = 4

(5) From three men and two women, an environment committee of two persons is to be formed. A: at least one woman member; B: one man, one woman; C: no woman member.

Let the men be M₁, M₂, M₃ and the women be W₁, W₂.

S = {M₁M₂, M₁M₃, M₂M₃, M₁W₁, M₁W₂, M₂W₁, M₂W₂, M₃W₁, M₃W₂, W₁W₂}

n(S) = 10

A = {M₁W₁, M₁W₂, M₂W₁, M₂W₂, M₃W₁, M₃W₂, W₁W₂}, n(A) = 7

B = {M₁W₁, M₁W₂, M₂W₁, M₂W₂, M₃W₁, M₃W₂}, n(B) = 6

C = {M₁M₂, M₁M₃, M₂M₃}, n(C) = 3

(6) One coin and one die are thrown simultaneously. A: head and an odd number; B: a head or tail and an even number; C: number greater than 7 and tail on the coin.

S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}, n(S) = 12

A = {H1, H3, H5}, n(A) = 3

B = {H2, H4, H6, T2, T4, T6}, n(B) = 6

A die has no number greater than 7, so C = { } = ∅, n(C) = 0

Practice Set 5.4

1If two coins are tossed, find the probability of the following events. (1) Getting at least one head. (2) Getting no head.

S = {HH, HT, TH, TT}, n(S) = 4

(1) At least one head

A = {HH, HT, TH}, n(A) = 3

P(A) = n(A)n(S) = 34

P(A) = 3/4

(2) No head

B = {TT}, n(B) = 1

P(B) = n(B)n(S) = 14

P(B) = 1/4

2If two dice are rolled simultaneously, find the probability of the following events.

n(S) = 36

(1) The sum of the digits on the upper faces is at least 10.

A = {(4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}, n(A) = 6

P(A) = n(A)n(S) = 636 = 16

P(A) = 1/6

(2) The sum of the digits on the upper faces is 33.

The greatest possible sum is 6 + 6 = 12, so B = { }, n(B) = 0

P(B) = n(B)n(S) = 036

P(B) = 0

(3) The digit on the first die is greater than the digit on second die.

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

n(C) = 15

P(C) = n(C)n(S) = 1536 = 512

P(C) = 5/12

3There are 15 tickets in a box, each bearing one of the numbers from 1 to 15. One ticket is drawn at random. Find the probability that the ticket drawn (1) shows an even number (2) shows a number which is a multiple of 5.

S = {1, 2, 3, …, 15}, n(S) = 15

(1) Even number

A = {2, 4, 6, 8, 10, 12, 14}, n(A) = 7

P(A) = n(A)n(S) = 715

P(A) = 7/15

(2) Multiple of 5

B = {5, 10, 15}, n(B) = 3

P(B) = n(B)n(S) = 315 = 15

P(B) = 1/5

4A two digit number is formed with digits 2, 3, 5, 7, 9 without repetition. What is the probability that the number formed is (1) an odd number (2) a multiple of 5?

S = {23, 25, 27, 29, 32, 35, 37, 39, 52, 53, 57, 59, 72, 73, 75, 79, 92, 93, 95, 97}

n(S) = 20

(1) Odd number

The unit digit must be 3, 5, 7 or 9, i.e. every number except 32, 52, 72, 92.

n(A) = 20 − 4 = 16

P(A) = n(A)n(S) = 1620 = 45

P(A) = 4/5

(2) Multiple of 5

B = {25, 35, 75, 95}, n(B) = 4

P(B) = n(B)n(S) = 420 = 15

P(B) = 1/5

5A card is drawn at random from a pack of well shuffled 52 playing cards. Find the probability that the card drawn is (1) an ace (2) a spade.

n(S) = 52

(1) An ace

There are 4 aces, so n(A) = 4

P(A) = n(A)n(S) = 452 = 113

P(A) = 1/13

(2) A spade

There are 13 spades, so n(B) = 13

P(B) = n(B)n(S) = 1352 = 14

P(B) = 1/4

Problem Set 5

1Choose the correct alternative answer for each of the following questions.

(1) Which number cannot represent a probability? (A) 23 (B) 1.5 (C) 15% (D) 0.7

A probability must lie from 0 to 1. 1.5 is greater than 1.

(B) 1.5

(2) A die is rolled. What is the probability that the number appearing on upper face is less than 3? (A) 16 (B) 13 (C) 12 (D) 0

A = {1, 2}, n(A) = 2

P(A) = 26 = 13

(B) 1/3

(3) What is the probability that a number chosen from 1 to 100 is a prime number? (A) 15 (B) 625 (C) 14 (D) 1350

There are 25 prime numbers from 1 to 100.

P = 25100 = 14

(4) There are 40 cards in a bag, each bearing a number from 1 to 40. What is the probability that the card drawn bears a multiple of 5? (A) 15 (B) 35 (C) 45 (D) 13

Multiples of 5: 5, 10, …, 40 → 8 numbers

P = 840 = 15

(A) 1/5

(5) If n(A) = 2, P(A) = 15, then n(S) = ? (A) 10 (B) 52 (C) 25 (D) 13

2n(S) = 15 ∴ n(S) = 10

(A) 10

2Basketball players John, Vasim, Akash were practising the ball drop in the basket. The probabilities of success for John, Vasim and Akash are 4/5, 0.83 and 58% respectively. Who had the greatest probability of success?

John: 45 = 0.80

Vasim: 0.83

Akash: 58% = 0.58

0.83 > 0.80 > 0.58

Vasim had the greatest probability of success.

3In a hockey team there are 6 defenders, 4 offenders and 1 goalee. One player is to be selected randomly as a captain. Find the probability that (1) the goalee will be selected (2) a defender will be selected.

n(S) = 6 + 4 + 1 = 11

(1) Goalee

n(A) = 1

P(A) = n(A)n(S) = 111

P(A) = 1/11

(2) Defender

n(B) = 6

P(B) = n(B)n(S) = 611

P(B) = 6/11

4Joseph kept 26 cards in a cap, bearing one English alphabet on each card. One card is drawn at random. What is the probability that the card drawn is a vowel card?

n(S) = 26

A = {a, e, i, o, u}, n(A) = 5

P(A) = n(A)n(S) = 526

P(A) = 5/26

5A balloon vendor has 2 red, 3 blue and 4 green balloons. He chooses one at random to give to Pranali. Find the probability that Pranali gets (1) a red balloon (2) a blue balloon (3) a green balloon.

n(S) = 2 + 3 + 4 = 9

(1) Red

P(A) = n(A)n(S) = 29

P(A) = 2/9

(2) Blue

P(B) = n(B)n(S) = 39 = 13

P(B) = 1/3

(3) Green

P(C) = n(C)n(S) = 49

P(C) = 4/9

6A box contains 5 red, 8 blue and 3 green pens. Rutuja wants to pick a pen at random. What is the probability that the pen is blue?

n(S) = 5 + 8 + 3 = 16

n(A) = 8

P(A) = n(A)n(S) = 816 = 12

P(A) = 1/2

7Six faces of a die are A, B, C, D, E, A. If the die is rolled once, find the probability that (1) ‘A’ appears on upper face (2) ‘D’ appears on upper face.

n(S) = 6

(1) ‘A’ appears

Two faces bear A, so n(A) = 2

P = 26 = 13

1/3

(2) ‘D’ appears

One face bears D, so n(D) = 1

P = 16

1/6

8A box contains 30 tickets, bearing only one number from 1 to 30 on each. If one ticket is drawn at random, find the probability that the ticket drawn bears (1) an odd number (2) a complete square number.

n(S) = 30

(1) Odd number

A = {1, 3, 5, …, 29}, n(A) = 15

P(A) = n(A)n(S) = 1530 = 12

P(A) = 1/2

(2) Complete square

B = {1, 4, 9, 16, 25}, n(B) = 5

P(B) = n(B)n(S) = 530 = 16

P(B) = 1/6

9Length and breadth of a rectangular garden are 77 m and 50 m. There is a circular lake in the garden having diameter 14 m. Due to wind, a towel from a terrace on a nearby building fell into the garden. Find the probability that it fell in the lake.

Here the sample space is the whole garden, so its area plays the role of n(S).

Area of garden = 77 × 50 = 3850 m²

Radius of lake = 14 ÷ 2 = 7 m

Area of lake = πr² = 227 × 7 × 7 = 154 m²

P(towel falls in the lake) = Area of lakeArea of garden = 1543850 = 125

Probability = 1/25

The towel is taken to be equally likely to land anywhere in the garden, since the question says it fell into the garden.

10In a game of chance, a spinning arrow comes to rest at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8. All these are equally likely outcomes. Find the probability that it will rest at (1) 8 (2) an odd number (3) a number greater than 2 (4) a number less than 9.

S = {1, 2, 3, 4, 5, 6, 7, 8}, n(S) = 8

(1) 8

A = {8}, n(A) = 1

P(A) = n(A)n(S) = 18

P(A) = 1/8

(2) An odd number

B = {1, 3, 5, 7}, n(B) = 4

P(B) = n(B)n(S) = 48 = 12

P(B) = 1/2

(3) A number greater than 2

C = {3, 4, 5, 6, 7, 8}, n(C) = 6

P(C) = n(C)n(S) = 68 = 34

P(C) = 3/4

(4) A number less than 9

D = {1, 2, 3, 4, 5, 6, 7, 8}, n(D) = 8

P(D) = n(D)n(S) = 88 = 1

P(D) = 1 (a sure event)

11There are six cards in a box, each bearing a number from 0 to 5. Find the probability that a card drawn shows (1) a natural number (2) a number less than 1 (3) a whole number (4) a number greater than 5.

S = {0, 1, 2, 3, 4, 5}, n(S) = 6

(1) A natural number

A = {1, 2, 3, 4, 5}, n(A) = 5

P(A) = n(A)n(S) = 56

P(A) = 5/6

(2) A number less than 1

B = {0}, n(B) = 1

P(B) = n(B)n(S) = 16

P(B) = 1/6

(3) A whole number

C = {0, 1, 2, 3, 4, 5}, n(C) = 6

P(C) = n(C)n(S) = 66 = 1

P(C) = 1

(4) A number greater than 5

D = { }, n(D) = 0

P(D) = n(D)n(S) = 06

P(D) = 0

12A bag contains 3 red, 3 white and 3 green balls. One ball is taken out at random. What is the probability that the ball drawn is (1) red (2) not red (3) either red or white?

n(S) = 3 + 3 + 3 = 9

(1) Red

n(A) = 3

P(A) = n(A)n(S) = 39 = 13

P(A) = 1/3

(2) Not red

White or green: n(B) = 3 + 3 = 6

P(B) = n(B)n(S) = 69 = 23

P(B) = 2/3

(3) Either red or white

n(C) = 3 + 3 = 6

P(C) = n(C)n(S) = 69 = 23

P(C) = 2/3

13Each card bears one letter from the word ‘mathematics’. The cards are placed on a table upside down. Find the probability that a card drawn bears the letter ‘m’.

The word ‘mathematics’ has 11 letters: m, a, t, h, e, m, a, t, i, c, s

n(S) = 11

The letter m appears 2 times, so n(A) = 2

P(A) = n(A)n(S) = 211

P(A) = 2/11

14Out of 200 students from a school, 135 like Kabaddi and the remaining students do not like the game. If one student is selected at random, find the probability that the student selected doesn’t like Kabaddi.

n(S) = 200

n(A) = 200 − 135 = 65

P(A) = n(A)n(S) = 65200 = 1340

P(A) = 13/40

15A two digit number is to be formed from the digits 0, 1, 2, 3, 4. Repetition of the digits is allowed. Find the probability that the number so formed is a (1) prime number (2) multiple of 4 (3) multiple of 11.

The tens digit can be 1, 2, 3 or 4 (not 0); the unit digit can be 0, 1, 2, 3 or 4.

S = {10, 11, 12, 13, 14,
20, 21, 22, 23, 24,
30, 31, 32, 33, 34,
40, 41, 42, 43, 44}

n(S) = 4 × 5 = 20

(1) Prime number

A = {11, 13, 23, 31, 41, 43}, n(A) = 6

P(A) = n(A)n(S) = 620 = 310

P(A) = 3/10

(2) Multiple of 4

B = {12, 20, 24, 32, 40, 44}, n(B) = 6

P(B) = n(B)n(S) = 620 = 310

P(B) = 3/10

(3) Multiple of 11

C = {11, 22, 33, 44}, n(C) = 4

P(C) = n(C)n(S) = 420 = 15

P(C) = 1/5

16The faces of a die bear numbers 0, 1, 2, 3, 4, 5. If the die is rolled twice, then find the probability that the product of digits on the upper face is zero.

n(S) = 6 × 6 = 36

The product is zero when at least one of the two numbers is 0.

A = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (1, 0), (2, 0), (3, 0), (4, 0), (5, 0)}

n(A) = 11

P(A) = n(A)n(S) = 1136

P(A) = 11/36

17Do the following activity.

Activity I: Students in your class wearing spectacles.

The numbers depend on your own class. Example: suppose there are 40 students and 8 of them wear spectacles.

n(S) = 40, n(A) = 8

P(A) = 840 = 15

Students not wearing spectacles: n(B) = 40 − 8 = 32, P(B) = 3240 = 45

Replace 40 and 8 with the actual numbers from your class. P(A) + P(B) will always be 1.

Activity II: Decide the sample space yourself. The condition for event A is ‘getting an even number’.

Example: one die is rolled.

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

A = {2, 4, 6}, n(A) = 3

∴ P(A) = 36 = 12

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