# Class 9 RD Sharma Solutions- Chapter 25 Probability – Exercise 25.1 | Set 1

### Question 1. A coin is tossed 1000 times with the following sequence:

### Head: 455, Tail: 545

### Compute the probability of each event.

**Solution:**

According to the given question,

Coin is tossed 1000 times, that means number of trials are 1000

Let us assume that, event of getting head and event

of getting tail be H and T respectively.

Number of trials in which the H happens = 455

Probability of H = (Number of heads) / (Total number of trials)

P(H) = 455/1000 = 0.455

Similarly,

Number of trials in which the T happens = 545

Probability of T = (Number of trials) / (Total number of trials)

Probability of the event getting a tail, P(T) = 545/1000 = 0.545

### Question 2: Two coins are tossed simultaneously 500 times with the following frequencies of different outcomes:

### Two heads: 95 times

### One tail: 290 times

### No head: 115 times

### Find the probability of occurrence of each of these events.

**Solution:**

According to the formula,

Probability of any event = (Number of favorable outcome) / (Total number of trials)Total number of trials = 95 + 290 + 115 = 500 -(given)

P(Getting two heads) = 95/500 = 0.19

P(Getting one tail) = 290/500 = 0.58

P(Getting no head) = 115/500 = 0.23

### Question 3: Three coins are tossed simultaneously 100 times with the following frequencies of different outcomes:

## Outcome | ## No Head | ## One Head | ## Two Head | ## Three Head |

## Frequency | ## 14 | ## 38 | ## 36 | ## 12 |

### If the three coins are simultaneously tossed again, compute the probability of:

### (i) 2 heads coming up

### (ii) 3 heads coming up

### (iii) At least one head coming up

### (iv) Getting more heads than tails

### (v) Getting more tails than heads

**Solution:**

According to the formula,

Probability of an event = (Number of Favorable outcomes) / (Total Numbers of outcomes)Given: Total numbers of outcomes = 100

(i)Probability of 2 Heads coming up = 36/100 = 0.36

(ii)Probability of 3 Heads coming up = 12/100 = 0.12

(iii)Probability of at least one head coming up = (38+36+12) / 100 = 86/100 = 0.86

(iv)Probability of getting more Heads than Tails = (36+12)/100 = 48/100 = 0.48

(v)Probability of getting more tails than heads = (14+38) / 100 = 52/100 = 0.52

### Question 4. 1500 families with 2 children were selected randomly, and the following data were recorded:

## No of the girls in the family | ## 0 | ## 1 | ## 2 |

## No of girls | ## 211 | ## 814 | ## 475 |

### If a family is chosen at random, compute the probability that it has:

### (i) No girl

### (ii) 1 girl

### (iii) 2 girls

### (iv) At most one girl

### (v) More girls than boys

**Solution:**

According to the formula,

Probability of an event = (Number of Favorable outcomes) / (Total Numbers of outcomes)Total numbers of outcomes = 211 + 814 + 475 = 1500

(Here, we are assuming that total numbers of outcomes = total number of families)

(i)Probability of having no girl = 211/1500 = 0.1406

(ii)Probability of having 1 girl = 814/1500 = 0.5426

(iii)Probability of having 2 girls = 475/1500 = 0.3166

(iv)Probability of having at the most one girl = (211+814) /1500 = 1025/1500 = 0.6833

(v)Probability of having more girls than boys = 475/1500 = 0.31

### Question 5. In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays. Find the probability that on a ball played:

### (i) He hits the boundary

### (ii) He does not hit a boundary.

**Solution:**

Total number of balls played by a player = 30 -(According to the question)

Number of times he hits a boundary = 6

Number of times he does not hit a boundary = 30 â€“ 6 = 24

According to the formula,

Probability of an event = (Number of Favorable outcomes) / (Total Numbers of outcomes)

(i)Probability (he hits boundary) = (Number of times he hit a boundary) / (Total number of balls he played)= 6/30 = 1/5

(ii)Probability that the batsman does not hit a boundary = 24/30 = 4/5

### Question 6. The percentage of marks obtained by a student in monthly unit tests are given below:

## Unit Test | ## I | ## II | ## III | ## IV | ## V |

## Percentage of Mark obtained | ## 69 | ## 71 | ## 73 | ## 68 | ## 76 |

### Find the probability that the student gets

### (i) More than 70% marks

### (ii) Less than 70% marks

### (iii) A distinction

**Solution:**

Total number of unit tests taken = 5

According to the formula,

Probability of an event = (Number of Favorable outcomes) / (Total Numbers of outcomes)

(i)Number of times student got more than 70% = 3Probability (Getting more than 70%) = 3/5 = 0.6

(ii)Number of times student got less than 70% = 2Probability (Getting less than 70%) = 2/5 = 0.4

(iii)Number of times student got a distinction = 1[Marks more than 75%]

Probability (Getting a distinction) = 1/5 = 0.2

### Question 7. To know the opinion of the students about Mathematics, a survey of 200 students were conducted. The data was recorded in the following table:

## Opinion | ## Like | ## Dislike |

## Number of students | ## 135 | ## 65 |

### Find the probability that student chosen at random:

### (i) Likes Mathematics

### (ii) Does not like it.

**Solution:**

Total number of students = 200

Students like mathematics = 135

Students dislike Mathematics = 65

According to the formula,

Probability of an event = (Number of Favorable outcomes) / (Total Numbers of outcomes)

(i)Probability (Student likes mathematics) = 135/200 = 0.675

(ii)Probability (Student does not like mathematics) = 65/200 = 0.325

### Question 8. The blood groups of 30 students of class ix are recorded as follows:

### A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O, A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O

### A student is selected at random from the class from blood donation. Find the probability that the blood groups of the student chosen is:

### (i) a (ii) b (iii) ab (iv) o

**Solution:**

(i)Probability of a student of blood group Favorable outcome ATotal outcome = (Number of Favorable outcomes) / (Total Numbers of outcomes)

= 9/30 = 0.3

(ii)Probability of a student of blood group Favorable outcome BTotal outcome = (Number of Favorable outcomes) / (Total Numbers of outcomes)

= 6/30 = 0.2

(iii)The probability of a student of blood groupProbability of an event = (Number of Favorable outcomes) / (Total Numbers of outcomes)

= 3/30 =0.1

(iv)The probability of a student of blood groupProbability of an event = (Number of Favorable outcomes) / (Total Numbers of outcomes)

= 12/30 = 0.4

### Question 9. 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.

**Solution:**

Given bag of wheat flour in the question

4.97, 5.05, 5.05, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00

The total number of wheat flour bags =11.

The number of wheat flour bags contain more than 5 Kg are 7.

Then probability of bags chosen at random =

(The number of wheat flour bags contain more than 5 Kg) / (The total number of wheat flour bags)

= 7/11

### Question 10. The following table shows the birth month of 40 students of class IX.

## Jan | ## Feb | ## March | ## April | ## May | ## June | ## July | ## Aug | ## Sept | ## Oct | ## Nov | ## Dec |

## 3 | ## 4 | ## 2 | ## 2 | ## 5 | ## 1 | ## 2 | ## 5 | ## 3 | ## 4 | ## 4 | ## 4 |

### Find the probability that a student was born in August.

**Solution:**

Probability (Students was born in August) Favorable outcome

Total outcome = (Favorable outcome)/(Total outcome)

= 6/40 = 3/20

## Please

Loginto comment...