Let E be the event of hitting the boundary.

Then,

∴ 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

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 =

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

having 1 girl

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

having no girl

Sum of these probabilities

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 =

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 =

(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 =

(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 =

(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 =

(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 =

542 Views