Updated on: 2026-03-31 | Author: Aarti Kulkarni

Algebra (Math 1) Chapter 5 Probability Solutions

Q1. A bag contains 5 white balls and some blue balls. If the probability of drawing a blue ball is double that of a white ball, determine the number of blue balls in the bag

Let the number of blue balls be x.

Number of white balls = 5
∴ Total number of balls = (x + 5)

P(blue ball is drawn) = `x/(x + 5)`

P(white ball is drawn) = `5/(x + 5)`

According to the given condition, the probability of drawing a blue ball is double that of a white ball.

∴ P(blue ball is drawn) = 2 × P(white ball is drawn)
∴ `x/(x +5) = 2 xx 5/(x + 5)`
∴ x(x + 5) = 10(x + 5)
∴ x2 + 5x = 10x + 50
∴ x2 – 5x – 50 = 0
∴ x2 – 10x + 5x – 50 = 0
∴ x(x – 10) + 5(x – 10) = 0
∴ (x – 10)(x + 5) = 0
∴ x – 10 = 0 or x + 5 = 0
∴ x = 10 or x = – 5

But, number of balls cannot be negative.

∴ x = 10
∴ The number of blue balls in the bag is 10.

Q2. A bag contains 8 red balls and some blue balls. If one ball is drawn randomly the probability of drawing a red ball to a blue ball are in the ratio 5 : 2, determine the probability of drawing a blue ball from the bag.

Let the number of red balls be 8.

Number of blue balls = x
∴ Total number of balls = (8 + x)
∴ `("P"("red ball is drawn"))/("P"("blue bal is draw")) = 5/2`

Let the probability of getting a red ball be A

Favourable number of outcome = 8
∴ p(A) = `8/(8 + x)`

Let probability of getting a blue ball be B

Favourable number of outcome = x
∴ p(B) = `x/(8 + x)`
∴ `("P"("red ball is drawn"))/("P"("blue bal is draw")) = 5/2`
`therefore (8/(8 + x))/(x/(8 + x)) = 5/2`
`=> 8/cancel(8 + x) xx cancel(8 + x)/x = 5/2`
`=> 8/x = 5/2`
∴ 5x = 16
∴ x = `16/5`
∴ P(B) = `x/(8 + x) = (16/5)/(8 + 16/5)`
`= (16/5)/(56/5) = cancel(16)^2/cancel(56)_7`

`= 2/7`

Q3. A balloon vendor has 2 red, 3 blue and 4 green balloons. He wants to choose one of them at random to give it to Pranali. What is the probability of the event that Pranali gets, a blue balloon.

Answer

Given: Number of red balloon = 2
Number of blue balloon = 3
Number of green balloon = 4

Let S be the sample space.

∴ n(S) = Number of red balloon + Number of blue balloon + Number of green balloon

= 2 + 3 + 4

= 9

Let B be the event of getting a blue balloon.

Given: n(B) = 3
∴ `"P"("B") = ("n"("B"))/("n"("S"))`

= `3/9`

= `1/3`

Hence, the probability of the event that Pranali gets a blue balloon is `1/3.`

Q4. A balloon vendor has 2 red, 3 blue and 4 green balloons. He wants to choose one of them at random to give it to Pranali. What is the probability of the event that Pranali gets, a red balloon

Answer

Given: Number of red balloon = 2
Number of blue balloon = 3
Number of green balloon = 4

Let S be the sample space.

∴ n(S) = Number of red balloon + Number of blue balloon + Number of green balloon

= 2 + 3 + 4

= 9

Let A be the event of getting a red balloon.

Given: n(A) = 2
∴ `P(A) = (n(A))/(n(S))`

= `2/9`

Hence, the probability of the event that Pranali gets a red balloon is `2/9`.

Q5. A box contains 36 cards, bearing only one number from 1 to 36 on each. If one card is drawn at random, find the probability of an event that the card drawn bears, a complete square number

Sample space,

S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10,

11, 12, 13, 14, 15, 16, 17, 18, 19, 20,

21, 22, 23, 24, 25, 26, 27, 28, 29, 30,

31, 32, 33, 34, 35, 36}

∴ n(S) = 36

Let A be the event that the card drawn bears a complete square number.

∴ A = {1, 4, 9, 16, 25, 36}
∴ n(A) = 6
∴ P(A) = `("n"("A"))/("n"("S"))`
∴ P(A) = `6/36`
∴ P(A) = `1/6`

Q6. A box contains 36 cards, bearing only one number from 1 to 36 on each. If one card is drawn at random, find the probability of an event that the card drawn bears, a number divisible by 3

Sample space,

S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10,

11, 12, 13, 14, 15, 16, 17, 18, 19, 20,

21, 22, 23, 24, 25, 26, 27, 28, 29, 30,

31, 32, 33, 34, 35, 36}

∴ n(S) = 36

Let C be the event that the card drawn bears a number divisible by 3.

∴ C = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36}
∴ n(C) = 12
∴ P(C) = `("n"("C"))/("n"("S"))`
∴ P(C) = `12/36`
∴ P(C) = `1/3`

Q7. A box contains 36 cards, bearing only one number from 1 to 36 on each. If one card is drawn at random, find the probability of an event that the card drawn bears, a prime number

Sample space,

S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10,

11, 12, 13, 14, 15, 16, 17, 18, 19, 20,

21, 22, 23, 24, 25, 26, 27, 28, 29, 30,

31, 32, 33, 34, 35, 36}

∴ n(S) = 36

Let B be the event that the card drawn bears a prime number.

∴ B = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31}
∴ n(B) = 11
∴ P(B) = `("n"("B"))/("n"("S"))`
∴ P(B) = `11/36`

Q8. A box contains 5 strawberry chocolates, 6 coffee chocolates and 2 peppermint chocolates. If one of the chocolates is picked from the box at random, find the probability of the following events by completing the activity. Activity : Let ‘S’ is the sample space. ∴ n(S) = 5 + 6 + 2 = 13 Event B: it is a peppermint chocolate. ∴ n(B) = `square` ∴ P(B) = `square/("n"("S"))` .......[Formula] P(B) = `square/13`

Let ‘S’ be the sample space.

∴ n(S) = 5 + 6 + 2 = 13

Event B: it is a peppermint chocolate.

∴ n(B) = 2
∴ P(B) = `("n"("B"))/("n"("S"))` .......[Formula]

P(B) = `2/13`

Q9. A box contains 5 strawberry chocolates, 6 coffee chocolates, and 2 peppermint chocolates. If one of the chocolates is picked from the box at random, find the probability of the following events by completing the activity. Event A: it is coffee chocolate. Activity: Let ‘S’ be the sample space. ∴ n(S) = 5 + 6 + 2 = 13 Event A: it is a coffee chocolate. ∴ n(A) = `square` ∴ P(A) = `square/("n"("S"))` .......[Formula] P (A) = `square/13`

Let ‘S’ be the sample space.

∴ n(S) = 5 + 6 + 2 = 13

Event A: it is a coffee chocolate.

∴ n(A) = 6
∴ P(A) = `("n"("A"))/("n"("S"))` .......[Formula]

P (A) = `6/13`

Q10. A card is drawn at random from a pack of well shuffled 52 playing cards. Find the probability that the card drawn is a spade.

There are 52 playing cards.

∴ n(S) = 52

Let B be the event that the card drawn is a spade.

There are 13 spade cards.

∴ n(B) = 13
∴ P(B) = `("n"("B"))/("n"("S"))`

= `13/52`

= `1/4`

Hence, the probability that the card drawn is a spade is =`1/4`.

Q11. A card is drawn from a well shuffled pack of 52 playing cards. Find the probability of the event, the card drawn is a red card. Activity: Let ‘S’ be the sample space. ∴ n(S) = 52 Event A: Card drawn is a red card. ∴ Total red cards = `square` hearts + 13 diamonds ∴ n(A) = `square` ∴ P(A) = `square/(n("S"))` ......[Formula] P(A) = `26/52` P(A) = `square`

Let ‘S’ be the sample space.

∴ n(S) = 52

Event A: Card drawn is a red card.

∴ Total red cards = 13 hearts + 13 diamonds
∴ n(A) = 26
∴ P(A) = `bb(n(A))/(n(S))` ......[Formula]

P(A) = `26/52`

P(A) = `bb(1/2)`

Q12. A missing helicopter is reported to have crashed somewhere in the rectangular region shown in the figure. What is the probability that it crashed inside the lake shown in the figure?

The helicopter is equally likely to crash anywhere in the region. In the figure, the length and the breadth of the rectangle are 9 m and 4.5 m respectively.

Area of the entire region where the helicopter can crash

= (9 × 4.5) m2 = 40.5 m2

Let A be the event that helicopter crashed inside the lake.

The lake is rectangular shaped.

Length of lake = 9 – 6 = 3 m
Breadth of lake = 4.5 – 2 = 2.5 m
Area of lake = length × breadth

= 3 × 2.5

= 7.5 m2

∴ P(A) = `("n"("A"))/("n"("S"))`

= `7.5/40.5`

= `75/405`

= `5/27`

The probability that the helicopter crashed inside the lake is `5/27`

Q13. A two-digit number is formed with digits 2, 3, 5 without repetition. Write the sample space.

Answer

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

n(S) = 6

Q14. Choose the correct alternative answer for the following question and write the alphabet. Which of the following number cannot represent a probability?

Answer

1.5

Q15. If n(A) = 5 , P(A) = `1/2` then n(S) = ?

10

Q16. If One coin and one die are thrown simultaneously, find the probability of the following events. Event A: To get a tail and an even number

Sample space,

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

(T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}

∴ n(S) = 12

Event A: To get a tail and an even number.

∴ A = {(T, 2), (T, 4), (T, 6)}
∴ n(A) = 3
∴ P(A) = `("n"("A"))/("n"("S"))`
∴ P(A) = `3/12`
∴ P(A) = `1/4`

Q17. If One coin and one die are thrown simultaneously, find the probability of the following events. Event B: To get head and an odd number

Sample space,

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

(T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}

∴ n(S) = 12

Event B: To get head and an odd number.

∴ B = {(H, 1), (H, 3), (H, 5)}
∴ n(B) = 3
∴ P(B) = `("n"("B"))/("n"("S"))`
∴ P(B) = `3/12`
∴ P(B) = `1/4`

Q18. If one die is rolled, then find the probability of event that the number on the upper face is greater than 6?

Sample space,

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

∴ n(S) = 6

Let A be the event that the number on the upper face is greater than 6.

Here, the greatest number is 6.

∴ Event A is an impossible event.
∴ A = { }
∴ n(A) = 0
∴ P(A) = `("n"("A"))/("n"("S"))`
∴ P(A) = `0/6` = 0

Q19. If one die is rolled, then find the probability of the following event by completing the activity. Event A: The number on the upper face is prime. Activity: Let ‘S’ be the sample space. S = {1, 2, 3, 4, 5, 6} ∴ n(S) = 6 Event A: Prime number on the upper face. A = {`square`} ∴ n(A) = 3 P(A) = `square/(n(S))` .....[Formula] = `square/6` ∴ P(A) = `1/square`

Let ‘S’ be the sample space.

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

∴ n(S) = 6

Event A: Prime number on the upper face.

A = {2, 3, 5}

∴ n(A) = 3
P(A) = `bb(n(A))/(n(S))` .....[Formula]

= `bb3/6`

∴ P(A) = `1/bb2`

Q20. If three coins are tossed simultaneously, find the probability of the event to get no head

Sample space,

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

∴ n(S) = 8

Let A be the event of getting no head.

∴ A = {TTT}
∴ n(A) = 1
∴ P(A) = `("n"("A"))/("n"("S"))`
∴ P(A) = `1/8`

Q21. If three coins are tossed simultaneously, find the probability of the following events Event A: To get no head.

Sample space,

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

∴ n(S) = 8

Let A be the event of getting no head.

∴ A = {TTT}
∴ n(A) = 1
∴ P(A) = `("n"("A"))/("n"("S"))`
∴ P(A) = `1/8`

Q22. If three coins are tossed simultaneously, find the probability of the following events. Event B: To get at least two heads.

Sample space,

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

∴ n(S) = 8

Let B be the event of getting at least two heads

∴ B = {HHH, HHT, HTH, THH}
∴ n(B) = 4
∴ P(B) = `("n"("B"))/("n"("S"))`
∴ P(B) = `4/8`
∴ P(B) = `1/2`

Q23. If two coins are tossed, find the probability of event getting head on both the coins

Sample space,

S = {HH, HT, TH, TT}

∴ n(S) = 4

Let A be the event of getting head on both the coins.

∴ A = {HH}
∴ n(A) = 1
∴ P(A) = `("n"("A"))/("n"("S"))`
∴ P(A) = `1/4`

Q24. If two dice are rolled simultaneously, find the probability of the following events. Event A: The sum of the digits on the upper faces is at least 10

Sample space,

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

Let A be the event that the sum of the digits on the upper faces is at least 10.

∴ A = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
∴ n(A) = 6
∴ P(A) = `("n"("A"))/("n"("S"))`

= `6/36`

∴ P(A) = `1/6`

Q25. If two dice are rolled simultaneously, find the probability of the following events. Event B: The sum of the digits on the upper faces is 33

Sample space,

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

Let B be the event that the sum of the digits on the upper faces is 33.

The sum of the digits on the upper faces can be maximum 12.

∴ Event B is an impossible event.
∴ B = { }
∴ n(B) = 0
∴ P(B) = `("n"("B"))/("n"("S"))`

= `0/36`

∴ P(B) = 0

Q26. In a set of 25 cards, each card bears only one number from 1 to 25. One card is drawn randomly. Write the sample space for this random experiment

Answer

S = {1, 2, 3, ...., 25}

Q27. In Adarsh High School, out of 30 students in a class 3 students wear glasses (spectacles). If a student in the class is randomly selected, find the probability that he or she wears glasses (spectacles) by completing the following activity Activity: There are a total of 30 students in the class. ∴ n(S) = `square` Event: A selected student wears glasses (spectacles). ∴ n(A) = `square` P(A) = `square/("n"("S"))` ......[Formula] P(A) = `square`

There are a total of 30 students in the class.

∴ n(S) = 30

Event A: selected student wears glasses (spectacles).

∴ n(A) = 3
P(A) = `("n"("A"))/("n"("S"))` ......[Formula]

P(A) = `3/30` = `1/10`

Q28. In how many ways a card can be drawn from a well shuffled pack of playing cards?

52

Q29. The 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.

Sample space,

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

(1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (1, 5),

(2, 0), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5),

(3, 0), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5),

(4, 0), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5),

(5, 0), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5)}

∴ n(S) = 36

Let A be the event that the product of digits on the upper face is zero.

∴ 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"))`
∴ P(A) = `11/36`

Q30. There are 30 cards in a box, each bearing one of the numbers from 1 to 30. One card is drawn at random from the box. Find the probability of event that the card drawn shows a number which is a multiple of 5

Sample space,

S = {1, 2, 3,…, 30}

∴ n(S) = 30

Let A be the event that the card drawn shows a number which is a multiple of 5.

∴ A = {5, 10, 15, 20, 25, 30}
∴ n(A) = 6
∴ P(A) = `("n"("A"))/("n"("S"))`
∴ P(A) = `6/30`
∴ P(A) = `1/5`

Q31. Three horses A, B and C are in a race, A is twice as likely to win as B and B is twice as likely to win as C. What are their probabilities of winning?

Let P(C) be x.

Probability of horse B winning the race is twice as likely as C.

∴ P(B) = 2 × P(C)

= 2x

Probability of horse A winning the race is twice as likely as B.

P(A) = 2 × P(B)

= 2 × 2x

= 4x

The total probability is 1.

∴ P(A) + P(B) + P(C) = 1
∴ 4x + 2x + x = 1
∴ 7x = 1
∴ x = `1/7`
∴ P(C) = x

= `1/7`

P(B) = 2x

= `2 xx 1/7`

= `2/7`

P(A) = 4x

= `4 xx 1/7`

= `4/7`

∴ P(A) = `4/7`

P(B) = `2/7`

P(C) = `1/7`

Q32. Two coins are tossed simultaneously. Write the sample space (S) and expected sample points in the given events by completing the activity. Event A: to get at least one head. Event B: to get no head. Activity: Let ‘S’ be the sample space, when two coins are tossed simultaneously. ∴ S = `{square, HT, TH, square}` i. Event A: to get at least one head. ∴ A = `{HH, square, TH}` ii. Event B: to get no head. ∴ B = `{square}`.

Answer

Activity: Let ‘S’ be the sample space, when two coins are tossed simultaneously.

∴ S = {HH, HT, TH, TT}

Event A: to get at least one head.

∴ A = {HH, HT, TH}

Event B: to get no head.

∴ B = {TT}.

Q33. What is the probability of the event that a number chosen from 1 to 50 is a prime number?

`3/10`

Explanation:

S = {1, 2, ..... , 50}

∴ n(S) = 50

Let A be the event of getting a prime number.

∴ A = {2, 3, 5, 7, 11, 13, 17 ,19, 23, 29, 31, 37, 41, 43, 47}
∴ n(A) = 15
∴ P(A) = `("n"("A"))/("n"("S")) = 15/50 = 3/10`

Q34. What is the probability that a leap year has 53 Sundays?

A leap year has 366 days i.e., 52 weeks and 2 days.

There are 52 Sundays in 52 weeks.

For the remaining 2 days,

Sample space is

S = {(Mon., Tue.), (Tue., Wed.), (Wed., Thur.),

(Thur., Fri.), (Fri., Sat.), (Sat., Sun.), (Sun., Mon.)}

∴ n(S) = 7

Let A be the event that the remaining two days are Sundays.

∴ A = {(Sat., Sun.), (Sun., Mon.)}
∴ n(A) = 2
∴ P(A) = `("n"("A"))/("n"("S"))`
∴ P(A) = `2/7`
∴ The probability that a leap year has 53 Sundays is `2/7`.

Q35. What is the probability that an ordinary year has 53 Sundays?

Ordinary year has 365 days

365 days = 52 weeks + 1 day

That 1 day may be Sun, Mon, Tue, Wed, Thu, Fri, Sat

Total no. of possible outcomes = 7

Let E ⟶ event of getting 53 Sundays

No. of favourable outcomes = 1 {Sun}

P(E) =`"No.of favorable outcomes"/"Total no.of possible outcomes"`=1/7

Q36. When a dice is thrown, the number of sample points in the sample space are ______

6

Q37. When two dice are thrown, the number of sample points in the sample space are ______

36

Q38. Which of the following options shows the highest probability?

0.83

Q39. Write a sample space if two coins are tossed simultaneously

Answer

When two coins are tossed,

S = {HH, HT, TH, TT}

Q40. Write a sample space when a die is thrown.

Answer

Sample space for one die thrown,

S = {1, 2, 3, 4, 5, 6}.

Q41. Write the event in the set form for the following random experiment. ‘If one die is thrown, the number obtained on the upper face is even.’

Answer

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

Let A be the event that the number obtained on the upper face is even.

∴ A = {2, 4, 6}

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