Important for :

1

C

Total number of ways of selecting 3 students from 25 students = ^{25}C_{3}

Number of ways of selecting 1 girl and 2 boys = selecting 2 boys from 15 boys and 1 girl from 10 girls

⇒ Number of ways in which this can be done =^{15}C_{2} × ^{10}C_{1}

Number of ways of selecting 1 girl and 2 boys = selecting 2 boys from 15 boys and 1 girl from 10 girls

⇒ Number of ways in which this can be done =

⇒ Required probability = |
(^{15}C_{2} × ^{10}C_{1}) |

(^{25}C_{3}) |

= | 21 |

46 |

Hence, option C is correct.

2

C

Since one die can be thrown in six ways to obtain any one of the six numbers marked on its six faces

⇒ Total number of elementary events = 6 x 6 x 6 = 216

Let A be the event of getting a total of at least 6.Then Ā denotes the event of getting a total of less than 6 i.e. 3, 4, 5.

⇒ Ā = { (1,1,1), (1,1,2), (1,2,1), (2,1,1), (1,1,3),(1,3,1), (3,1,1), (1,2,2), (2,1,2), (2,2,1) }

So, favorable number of cases = 10

⇒ Total number of elementary events = 6 x 6 x 6 = 216

Let A be the event of getting a total of at least 6.Then Ā denotes the event of getting a total of less than 6 i.e. 3, 4, 5.

⇒ Ā = { (1,1,1), (1,1,2), (1,2,1), (2,1,1), (1,1,3),(1,3,1), (3,1,1), (1,2,2), (2,1,2), (2,2,1) }

So, favorable number of cases = 10

⇒ P(Ā) = | 10 |

216 |

⇒ 1 – P (A) = | 10 |

216 |

⇒ P(A) = 1 – | 10 |

216 |

= | 103 |

108 |

Hence, option (C) is correct.

3

4

B

Let rotten apples = 10 × | 2 | = 4, others = 6 |

5 |

If 1 apple is rotten + 2 apples are other

=

If 2 apples are rotten + 1 apple is other

=

If 3 apples are rotten

=

Total outcomes =

Probability = | 60 + 36 + 4 |

120 |

= | 100 | = | 5 |

120 | 6 |

Hence, option B is correct.

5