Probability -

EXERCISE-15.1

P1. In a cricket math, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.

Sol. Number of times the batswoman hits a boundary = 6

Total number of balls played = 30

$∴$ Number of times that the batswoman does not hit a boundary = 30 − 6 = 24

P (She does not hit a boundary) = $\frac{N u m b e r o f t i m e s w h e n s h e d o e s n o t h i t b o u n d a r}{T o t a l n u m b e r o f b a l l s p l a y e s}$

$= \frac{24}{30} = \frac{4}{5}$

P2. 1500 families with 2 children were selected randomly, and the following data were recorded :

Number of girls in a family	2	1	0
Number of families	475	814	211

Compute the probability of a family, chosen at random, having

(i) 2 girls

(ii) 1 girl

(iii) No girl Also check whether the sum of these probabilities is 1.

Sol. Total number of families = 475 + 814 + 211 = 1500

(i) Number of families having 2 girls = 475

P₁(a randomly chosen family has 2 girls) = $\frac{N u m b e r o f f a m i l i e s h a v i n g 2 g i r l s}{T o t a l n u m b e r o f f a m i l i e s}$

$= \frac{475}{1500} = \frac{19}{60}$

(ii) Number of families having 1 girl = 814

P₂ (a randomly chosen family has 1 girl) = $\frac{N u m b e r o f f a m i l i e s h a v i n g 1 g i r l s}{T o t a l n u m b e r o f f a m i l i e s}$

$= \frac{814}{1500} = \frac{407}{750}$

(iii) Number of families having no girl = 211

P₃ (a randomly chosen family has no girl) = $\frac{N u m b e r o f f a m i l i e s h a v i n g n o g i r l s}{T o t a l n u m b e r o f f a m i l i e s} = \frac{211}{1500}$

Sum of all these probabilities $= \frac{19}{60} + \frac{407}{750} + \frac{211}{1500}$

$= \frac{475 + 814 + 211}{1500} = \frac{1500}{1500} = 1$

Therefore, the sum of all these probabilities is 1.

P3. In a particular section of Class IX, 40 students were asked about the months of their birth and the following graph was prepared for the data so obtained :

Find the probability that a student of the class was born in August.

Sol. Number of students born in the month of August = 6

Total number of students = 40

P (Students born in the month of August) = $\frac{N u m b e r o f s t u d e n t b o r n i n A u g u s t}{T o t a l n u m b e r o f s t u d e n t} = \frac{6}{40} = \frac{3}{20}$

P4. Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:

Outcome	3 heads	2 heads	1 head	No head
Frequency	23	72	77	28

If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.

Sol. Number of times 2 heads come up = 72

Total number of times the coins were tossed = 200

P ( 2 heads will come up) = $\frac{N u m b e r o f t i m e s 2 h e a d s c o m e u p}{T o t a l n u m b e r o f t i m e s t h e c o i n w e r e t o s s e d} = \frac{72}{200} = \frac{9}{2}$

P5. An organization selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below:

MONTHLY INCOME (in Rs)	Vehicles per family
MONTHLY INCOME (in Rs)	0	1	2	Above 2
Less than 7000	10	160	25	0
7000 − 10000	0	305	27	2
10000 − 13000	1	535	29	1
13000 − 16000	2	469	59	25
16000 or more	1	579	82	88

Suppose a family is chosen, find the probability that the family chosen is

(i) earning Rs 10000 − 13000 per month and owning exactly 2 vehicles.

(ii) earning Rs 16000 or more per month and owning exactly 1 vehicle.

(iii) earning less than Rs 7000 per month and does not own any vehicle.

(iv) earning Rs 13000 − 16000 per month and owning more than 2 vehicles. (v) owning not more than 1 vehicle.

Sol. Number of total families surveyed = 10 + 160 + 25 + 0 + 0 + 305 + 27 + 2 + 1 + 535 + 29 + 1 + 2 + 469 + 59 + 25 + 1 + 579 + 82 + 88 = 2400

(i) Number of families earning Rs 10000 − 13000 per month and owning exactly 2 vehicles = 29 Hence, required probability, P = $\frac{29}{2400}$

(ii) Number of families earning Rs 16000 or more per month and owning exactly 1 vehicle = 579 Hence, required probability, $P = \frac{579}{2400}$

(iii) Number of families earning less than Rs 7000 per month and does not own any vehicle = 10 Hence, required probability, $P = \frac{10}{2400} = \frac{1}{240}$

(iv) Number of families earning Rs 13000 − 16000 per month and owning more than 2 vehicles = 25 Hence, required probability, $P = \frac{25}{2400} = \frac{1}{96}$

(v) Number of families owning not more than 1 vehicle = 10 + 160 + 0 + 305 + 1 + 535 + 2 + 469 + 1 + 579 = 2062

Hence, required probability, $P = \frac{2062}{2400} = \frac{1031}{1200}$

P6. A teacher wanted to analyse the performance of two sections of students in a mathematics test of 100 marks. Looking at their performances, she found that a few students got under 20 marks and a few got 70 marks or above. So she decided to group them into intervals of varying sizes as follows: 0 − 20, 20 − 30… 60 − 70, 70 − 100. Then she formed the following table :

(i) Find the probability that a student obtained less than 20 % in the mathematics test.

(ii) Find the probability that a student obtained marks 60 or above.

Sol. Total number of students = 90

(i) Number of students getting less than 20 % marks in the test = 7

Hence, required probability, $P = \frac{7}{90}$

(ii) Number of students obtaining marks 60 or above = 15 + 8 = 23

Hence, required probability, P = $\frac{23}{90}$

P7. To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table.

Opinion

Number of students

like

dislike

135

Find the probability that a student chosen at random

(i) likes statistics,

(ii) does not like it

Sol. Total number of students = 135 + 65 = 200

(i) Number of students liking statistics = 135

P (students liking statistics) = $\frac{135}{200} = \frac{27}{40}$

(ii) Number of students who do not like statistics = 65

P (students not liking statistics) = $\frac{65}{200} = \frac{13}{40}$

P8. The distance (in km) of 40 engineers from their residence to their place of work were found as follows.

5	3	10	20	25	11	13	7	12	31
19	10	12	17	18	11	32	17	16	2
7	9	7	8	3	5	12	15	18	3
12	14	2	9	6	15	15	7	6	12

What is the empirical probability that an engineer lives :

(i) less than 7 km from her place of work?

(ii) more than or equal to 7 km from her place of work?

(iii) within $\frac{1}{2}$ km from her place of work ?

Sol. (i) Total number of engineers = 40

Number of engineers living less than 7 km from their place of work = 9

Hence, required probability that an engineer lives less than 7 km from her place of work, $P = \frac{9}{40}$

(ii) Number of engineers living more than or equal to 7 km from their place of work = 40 − 9 = 31

Hence, required probability that an engineer lives more than or equal to 7 km from her place of work, $P = \frac{31}{40}$

(iii) Number of engineers living within $\frac{1}{2}$ km from her place of work = 0

Hence, required probability that an engineer lives within $\frac{1}{2}$ km from her place of work, P = 0

P11. 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.

Sol. Number of total bags = 11

Number of bags containing more than 5 kg of flour = 7

Hence, required probability, $P = \frac{7}{11}$

P12.

Concentration of SO₂ (in ppm)	Number of days (frequency )
0.00 − 0.04	4
0.04 − 0.08	9
0.08 − 0.12	9
0.12 − 0.16	2
0.16 − 0.20	4
0.20 − 0.24	2
Total	30

The above frequency distribution table represents the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12 − 0.16 on any of these days.

Sol. Number days for which the concentration of sulphur dioxide was in the interval of 0.12 − 0.16 = 2

Total number of days = 30

Hence, required probability, $P = \frac{2}{30} = \frac{1}{15}$

P 13:

Blood group	Number of students
A	9
B	6
AB	3
O	12
Total	30

The above frequency distribution table represents the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group AB.

Sol. Number of students having blood group AB = 3

Total number of students = 30

Hence, required probability, $P = \frac{3}{30} = \frac{1}{10}$

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