Chances and probability in real world problems worksheets, solved worksheet problems or exercises with step-by-step work, formative assessment as online test, calculator and more learning resources to learn, practice, assess and master the basic math skills of statistics.

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1

When a dice is rolled, find the probability to get the number which is greater than 3?

2

What is the probability of getting a king from a deck of cards?

3

A coin is tossed 1000 times with the following frequencies (i)Head:780 (ii)Tail:220. compute the probability for each event.

4

Two unbiased coins are tossed simultaneously find the probability of getting (i)two heads (ii)one head (iii)at least one head (iv)at most one head.

5

Two dice are thrown simultaneously. Find the probability of getting (i)sum is equal to 4 (ii)an even number as the sum.

6

A bag contains 10 red and 8 blue balls. One ball is drawn at random. Find the probability that the ball drawn is blue.

7

1500 families were surveyed and following data was recorded about their maids at homes. A family is selected at random. Find the probability that the family selected has (i)Both types of maids (ii)Part time maids (iii)No maids

Types of maids | Only part time | Only full time | Both |

Number of families | 860 | 370 | 250 |

8

1500 families with 2 children were selected randomly, and the following data were recorded.Compute the probability of a family, chosen at random, having (i)2 girls (ii)1 girl (iii)No girl.

Number of girls in a family | 2 | 1 | 0 |

number of families | 475 | 814 | 211 |

9

If a probability of a player winning a particular tennis match is 0.72. What is the probability of the player loosing the match?

10

A company manufactures 1000 Computers in 6 months. Out of which 38 of them are found to be defective. When you choose one Computer from the manufactured, what is the probability that selected Computer is a good one.

11

The record of a weather station shows that out of the past 300 consecutive days, its weather was forecasted correctly 195 times. What is the probability that on a given day selected at random, (i)it was correct (ii)it was not correct.

12

There are 24 balls in a pot. If 3 of them are Red, 5 of them are Blue and the remaining are Green then, what is the probability of picking out (i)a Blue ball, (ii)a Red ball and (iii)a Green ball?

13

In a football match, a goalkeeper of a team can stop the goal, 32 times out of 40 attempts tried by a team. Find the probability that the opponent team can convert the attempt into a goal.

14

Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg). 4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00. Find the probability that any of these bags chosen at random contains more than 5 kg of flour.

Answers Key

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1

When a dice is rolled, find the probability to get the number which is greater than 3?

Probability to get the number which is greater than 3 is 1/2

Solution

Sample space for a dice, S= {1, 2, 3, 4, 5, 6}

Let E be the event of getting a number greater than 3

E = {4,5,6}

P(E) = n(E)/n(S)

where n(E) = Number of Successful Events,

n(S) = Total Events of Sample Space

n(S) = Total Events of Sample Space

P(E) = 3/6

∴ Probability to get the number which is greater than 3= 1/2

2

What is the probability of getting a king from a deck of cards?

∴ Probability of getting a king from a deck of cards is 1/13

Solution

Sample space for a deck of cards, S = {spade (Ace, 2, 3, 4, 5, 6, 7, 8, 9,10, Jack, Queen, King), clubs (Ace, 2, 3, 4, 5, 6, 7, 8, 9,10, Jack, Queen, King), heart (Ace, 2, 3, 4, 5, 6, 7, 8, 9,10, Jack, Queen, King), diamond (Ace, 2, 3, 4, 5, 6, 7, 8, 9,10, Jack, Queen, King)}

Let E be the event of getting a king from a deck of cards

E = {spade king, clubs king, heart king, diamond king}

P(E) = n(E)/n(S)

where n(E) = Number of Successful Events,

n(S) = Total Events of Sample Space

n(S) = Total Events of Sample Space

P(E) = 4/52

∴ Probability of getting a king from a deck of cards= 1/13

3

A coin is tossed 1000 times with the following frequencies (i)Head:780 (ii)Tail:220. compute the probability for each event.

(i) Probability of getting a head is 39/50

(ii) Probability of getting a tail is 11/50

(ii) Probability of getting a tail is 11/50

Solution

**(i)** P(H) = n(H)/n(S)

**(ii)** P(T) = n(T)/n(S)

__Result__

(i) Probability of getting a head is 39/50

(ii) Probability of getting a tail is 11/50

Total number of trials, n(S) = 1000

Number of heads, n(H) = 780

Number of tails, n(T) = 220

where n(H) = Number of heads,

n(S) = Total number of trials

n(S) = Total number of trials

P(H) = 780/1000

∴ Probability of getting a head, P(H)= 39/50

where n(T) = Number of tails,

n(S) = Total number of trials

n(S) = Total number of trials

P(T) = 220/1000

∴ Probability of getting a tail, P(T)= 11/50

(i) Probability of getting a head is 39/50

(ii) Probability of getting a tail is 11/50

4

Two unbiased coins are tossed simultaneously find the probability of getting (i)two heads (ii)one head (iii)at least one head (iv)at most one head.

(i) Probability of getting two heads is 1/4

(ii) Probability of getting one head is 1/2

(iii) Probability of getting at least one head is 3/4

(iv) Probability of getting at most one head is 3/4

(ii) Probability of getting one head is 1/2

(iii) Probability of getting at least one head is 3/4

(iv) Probability of getting at most one head is 3/4

Solution

**(i) ** Probability of getting two heads, P(A)= n(A)/n(S)

**(ii) ** Probability of getting one head, P(B)= n(B)/n(S)

**(iii) ** Probability of getting at least one head, P(C)= n(C)/n(S)

**(iv) ** Probability of getting at most one head, P(D)= n(D)/n(S)

__Result__

(i) Probability of getting two heads is 1/4

(ii) Probability of getting one head is 1/2

(iii) Probability of getting at least one head is 3/4

(iv) Probability of getting at most one head is 3/4

Sample space when two coins are tossed (S) = {HH, TT, HT, TH}

Event of getting two heads, (A)= {HH}

Event of getting one head, (B)= {HT,TH}

Event of getting at least one head, (C)= {HH,TH,HT}

Event of getting at most one head, (D)= {HH,HT,TH}

P(A) = 1/4

P(B) = 2/4

P(C) = 3/4

P(D) = 3/4

(i) Probability of getting two heads is 1/4

(ii) Probability of getting one head is 1/2

(iii) Probability of getting at least one head is 3/4

(iv) Probability of getting at most one head is 3/4

5

Two dice are thrown simultaneously. Find the probability of getting (i)sum is equal to 4 (ii)an even number as the sum.

(i)Probability of getting sum is equal to 4 is 1/12

(ii) Probability of getting even number as the sum is 1/2

(ii) Probability of getting even number as the sum is 1/2

Solution

**(i) **Event of getting sum is equal to 4, (E) are given below, (1,3) (2,2) (3,1)

**(ii) **Event of getting even number as the sum, (E) are given below,

__Result__

(i) Probability of getting sum is equal to 4 is 1/12

(ii) Probability of getting even number as the sum is 1/2

Sample space for two dice, S = 36

Possible outcomes of two dice are given below,

(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)

P(E) = n(E)/n(S)

where n(E) = Number of Successful Events,

n(S) = Number of Successful Events

n(S) = Number of Successful Events

P(E) = 3/36

∴ Probability of getting sum is equal to 4, P(E)= 1/12

(1,1), (1,3), (1,5)

(2,2), (2,4), (2,6)

(3,1), (3,3), (3,5)

(4,2), (4,4), (4,6)

(5,1), (5,3), (5,5)

(6,2), (6,4), (6,6)

P(E) = n(E)/n(S)

where n(E) = Number of Successful Events,

n(S) = Number of Successful Events

n(S) = Number of Successful Events

P(E) = 18/36

∴ Probability of getting even number as the sum, P(E)= 1/2

(i) Probability of getting sum is equal to 4 is 1/12

(ii) Probability of getting even number as the sum is 1/2

6

A bag contains 10 red and 8 blue balls. One ball is drawn at random. Find the probability that the ball drawn is blue.

Probability of getting blue ball is 4/9

Solution

Total number of balls, n(S)= 18

Event of getting red ball, n(R)= 10

Event of getting blue ball, n(B)= 8

Probability of getting blue ball, P(B)= n(B)/n(S)

P(B)= 8/18

Probability, P(B)= 4/9

∴ Probability of getting blue ball is 4/9

∴ Probability of getting blue ball is 4/9

7

1500 families were surveyed and following data was recorded about their maids at homes. A family is selected at random. Find the probability that the family selected has (i)Both types of maids (ii)Part time maids (iii)No maids

Types of maids | Only part time | Only full time | Both |

Number of families | 860 | 370 | 250 |

(i)Probability that family selected has both types of maids is 1/6

(ii) Probability that family selected has part time maids is 43/75

(iii) Probability that family selected has no maids is 1/75

(ii) Probability that family selected has part time maids is 43/75

(iii) Probability that family selected has no maids is 1/75

Solution

**(i) **Probability that family selected have both types of maids, P(B)= n(B)/n(S)

**(ii) **Probability that family selected have part time maids, P(P)= n(P)/n(S)

**(iii) ** Family selected have maids= n(P)+n(F)+n(B)

__Result__

(i) Probability that family selected has both types of maids is 1/6

(ii) Probability that family selected has part time maids is 43/75

(iii) Probability that family selected has no maids is 1/75

Total families, n(S)= 1500

Only part time maids, n(P)= 860

Only full time maids, n(F)= 370

Both, n(B) = 250

P(B)= 250/1500

Probability, P(B)= 1/6

P(P)= 860/1500

Probability, P(P)= 43/75

∴ Family selected have maids= 860+370+250 = 1480

Family selected have no maids, n(N)= 1500-1480 = 20

Probability that family selected have no maids, P(N)= n(N)/n(S)

P(N)= 20/1500

Probability, P(N)= 1/75

(i) Probability that family selected has both types of maids is 1/6

(ii) Probability that family selected has part time maids is 43/75

(iii) Probability that family selected has no maids is 1/75

8

1500 families with 2 children were selected randomly, and the following data were recorded.Compute the probability of a family, chosen at random, having (i)2 girls (ii)1 girl (iii)No girl.

Number of girls in a family | 2 | 1 | 0 |

number of families | 475 | 814 | 211 |

(i) Probability of a family having 2 girls is 19/60

(ii) Probability of a family having 1 girl is 407/750

(iii) Probability of a family having no girl is 211/1500

(ii) Probability of a family having 1 girl is 407/750

(iii) Probability of a family having no girl is 211/1500

Solution

**(i) **Probability of a family having 2 girls, P(A)= n(A)/n(S)

**(ii) **Probability of a family having 1 girl, P(B)= n(B)/n(S)

**(iii) **Probability of a family having no girls, P(C)= n(C)/n(S)

__Result__

(i) Probability of a family having 2 girls is 19/60

(ii) Probability of a family having 1 girl is 407/750

(iii) Probability of a family having no girl is 211/1500

Total families, n(S) = 1500

Numbers of families having 2 girls, n(A)= 475

Numbers of families having 1 girl, n(B)= 814

Numbers of families having no girl, n(C)= 211

P(A) = 475/1500

Probability, P(A)= 19/60

P(B) = 814/1500

Probability, P(B)= 407/750

P(C) = 211/1500

Probability, P(C)= 211/1500

(i) Probability of a family having 2 girls is 19/60

(ii) Probability of a family having 1 girl is 407/750

(iii) Probability of a family having no girl is 211/1500

9

If a probability of a player winning a particular tennis match is 0.72. What is the probability of the player loosing the match?

Probability that it will not rain tomorrow is 0.28

Solution

Let E be the event that it will rain tomorrow

and E^{`} be the event that it will not rain tomorrow

P(E) = 0.72

P(E^{`}) = 1-0.72 = 0.28

∴ Probability that it will not rain tomorrow= 0.28

10

A company manufactures 1000 Computers in 6 months. Out of which 38 of them are found to be defective. When you choose one Computer from the manufactured, what is the probability that selected Computer is a good one.

Probability that selected Computer is a good one is 481/500

Solution

Total computers, n(S)= 1000

Defective computers, n(D)= 38

Number of good computers, n(G)= 1000-38 = 962

Probability that selected Computer is a good one, P(G)= n(G)/n(S)

P(G) = 962/1000

Probability, P(G)= 481/500

∴ Probability that selected Computer is a good one is 481/500

11

The record of a weather station shows that out of the past 300 consecutive days, its weather was forecasted correctly 195 times. What is the probability that on a given day selected at random, (i)it was correct (ii)it was not correct.

(i) The number of days when the forecast was correct is 13/20

(ii) The number of days when the forecast was not correct is 7/20

(ii) The number of days when the forecast was not correct is 7/20

Solution

**(i) **The number of days when the forecast was correct, n(C)= 195

**(ii) **The number of days when the forecast was not correct, n(N)= n(S)-n(C)

__Result__

(i) The number of days when the forecast was correct is 13/20

(ii) The number of days when the forecast was not correct is 7/20

Total number of days, n(S)= 300

Probability that selected day selected was correct, P(C)= n(C)/n(S)

P(C)= 195/300

Probability, P(C) = 13/20

∴ n(N) = 300-195 =105

Probability that selected day selected was not correct, P(N)= n(N)/n(S)

P(N) = 105/300

Probability, P(N)= 7/20

(i) The number of days when the forecast was correct is 13/20

(ii) The number of days when the forecast was not correct is 7/20

12

There are 24 balls in a pot. If 3 of them are Red, 5 of them are Blue and the remaining are Green then, what is the probability of picking out (i)a Blue ball, (ii)a Red ball and (iii)a Green ball?

(i) Probability of picking out blue ball is 5/24

(ii) Probability of picking out red ball is 1/8

(iii) Probability of picking out green ball is 2/3

(ii) Probability of picking out red ball is 1/8

(iii) Probability of picking out green ball is 2/3

Solution

**(i) **Probability of picking out blue ball, P(B)= n(B)/n(S)

**(ii) **Probability of picking out red ball, P(R)= n(R)/n(S)

**(iii) **Number of green ball, n(G)= n(S)-n(R)+n(B)

__Result__

(i) Probability of picking out blue ball is 5/24

(ii) Probability of picking out red ball is 1/8

(iii) Probability of picking out green ball is 2/3

Total number of balls, n(S)= 24

Number of red balls, n(R)= 3

Number of blue balls, n(B)= 5

P(B)= 5/24

Probability, P(B)= 5/24

P(R)= 3/24

Probability, P(R)= 1/8

n(G)= 24-(3+5)

n(G)= 16

Probability of picking out green ball, P(G)= n(G)/n(S)

P(G)= 16/24

Probability, P(G)= 2/3

(i) Probability of picking out blue ball is 5/24

(ii) Probability of picking out red ball is 1/8

(iii) Probability of picking out green ball is 2/3

13

In a football match, a goalkeeper of a team can stop the goal, 32 times out of 40 attempts tried by a team. Find the probability that the opponent team can convert the attempt into a goal.

Probability of attempts by opponent team is 1/5

Solution

Total number of attempts, n(S)= 40

Number of attempts by a team, n(A)= 32

Number of attempts by opponent team, n(B)= n(S)-n(A)

n(B)= 40-32

n(B)= 8

Probability of attempts by opponent team, P(B)= n(B)/n(S)

P(B)= 8/40

Probability, P(B)= 1/5

∴ Probability of attempts by opponent team is 1/5

14

Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg). 4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00. Find the probability that any of these bags chosen at random contains more than 5 kg of flour.

Probability that bag chosen at random contains more than 5 kg of flour is 7/11

Solution

Total number of bags present, n(S)= 11

Number of bags containing more than 5 kg of flour, n(A)= 7

Probability that any of these bags chosen at random contains more than 5 kg of flour, P(A)= n(A)/n(S)

P(A)= 7/11

Probability, P(A)= 7/11

∴ Probability that bag chosen at random contains more than 5 kg of flour is 7/11

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