zigya tab

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


Let E be the event of hitting the boundary.

Then,

straight P left parenthesis straight E right parenthesis equals fraction numerator his space the space boundary over denominator Total end fraction equals 6 over 30 equals 1 half equals 0.2

∴ Probability of not hitting the boundary
= 1 – Probability of hitting the boundary
= 1 – P(E) = 1 – 0.2 = 0.8.



1008 Views

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

71

28

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


Total number of times the three coins are tossed = 200
Number of times when 2 heads appear = 72
∴ Probability of 2 heads coming up

equals 72 over 200 equals 9 over 25.

1243 Views

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

(0 2 girls (ii) 1 girl (iii) No girl.

Also check whether the sum of these probabilities is 1.


Total number of families
= 475 + 814 + 211 = 1500
(i) Probability of a family, chosen at random,

having 2 girls = 475 over 1500 equals 19 over 60

(ii) Probability of a family, chosen at random,

having 1 girl  equals 814 over 1500 equals 407 over 750

(iii) Probability of a family, chosen at random,

having no girl  equals 211 over 1500

Sum of these probabilities

equals 19 over 60 plus 407 over 750 plus 211 over 1500
equals fraction numerator 475 plus 814 plus 211 over denominator 1500 end fraction equals 1500 over 1500 equals 1

Hence, the sum is checked.


958 Views

In a particular section of Class IX, 40 students were asked about the months of their birth, the following graph was prepared for the data so obtained. Find the probability that a student of the class was bom in August.


Total number of students bom in the year =3+4+2+2+5+1+2+6+3+4+4+4=40
Number of students bom in August = 6
∴ Probability that a student of the class was

born in August = 6 over 40 equals 3 over 20.

373 Views

An organisation 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

Vehicles per family

(in र)

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  र 10000–13000 per month and owning exactly 2 vehicles.
(ii) earning र 16000 or more per month and owning exactly l vehicle.
(iii) earning less than र 7000 per month and does not own any vehicle.
(iv) earning र 13000–16000 per month and owning more than 2 vehicles.
(v) owning not more than I vehicle.


Total number of families selected = 2400
(i) Number of families earning र10000–13000 per month and owning exactly 2 vehicles = 29
∴ Probability that the family chosen is earning र10000–13000 per month and owning

exactly 2 vehicles = 29 over 2400.

(ii) Number of families earning र16000 or more per month and owning exactly 1 vehicle = 579

∴ Probability that the family chosen is earning र16000 or more per month and owning

exactly 1 vehicle = 579 over 2400 equals 193 over 800.

(iii) Number of families earning less than र 7000 per month and does not own any vehicle =10

Probability that the family chosen is earning less than र7000 per month and does not own any vehicle = 10 over 2400 equals 1 over 240
(iv) Number of families earning र13000–16000 per month and owning more than 2 vehicles = 25

∴ Probability that the family chosen is earning र13000–16000 per month and owning
more than 2 vehicles = 25 over 2400 equals 1 over 96.
(v) Number of families owning not more than 1 vehicle

= Number of families owning 0 vehicle + Number of families owning 1 vehicle = (10 + 0+ 1 + 2 + 1) +(160 + 305 + 535 + 469 +579) = 14 + 2048 = 2062
∴ Probability that the family chosen owns not more than 1 vehicle = 2062 over 2400 equals 1031 over 1200




542 Views