A) 1/3 | B) 3/5 |

C) 2/3 | D) 5/6 |

Explanation:

S = { 1, 2, 3, 4, .....18 }

=> n(S) = 18

E1 = {2, 4, 6, 8, 10, 12, 14, 16, 18}

=> n(E1) = 9

E2 = {3, 6, 9, 12, 15, 18 }

=> n(E2) = 6

=> n(E3) = 3

=> n(E) = 9 + 6 - 3 =12

where E = { 2, 3, 4, 6, 8, 9, 10, 12, 12, 14, 15, 16, 18 }

A) 1 | B) 1/2 |

C) 0 | D) 3/5 |

Explanation:

The number of exhaustive outcomes is 36.

Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then = 18/36 = 1/2

P(E) = 1 - 1/2 = 1/2.

A) 1/10 | B) 1/9 |

C) 1 | D) 0 |

Explanation:

As we know we have 10 letter and 10 different address and one more information given that exactly 9 letter will at the correct address....so the remaining one letter automatically reach to their correct address

P(E) = favorable outcomes /total outcomes

Here favorable outcomes are '0'.

So probability is '0'.

A) [1, 32] | B) (0, 1) |

C) [1, 1/2] | D) (1, 1/2] |

Explanation:

Probability of atleast one failure

= 1 - no failure 31/32

= 1 - 31/32

= 1/32

= p 1/2

Also p 0

Hence p lies in [0,1/2].

A) 1/3 | B) 1/2 |

C) 2/3 | D) 2/5 |

Explanation:

In a family with 2 children there are four possibilities:

1) the first child is a boy and the second child is a boy (bb)

2) the first child is a boy and the second child is a girl (bg)

3) the first child is a girl and the second child is a boy (gb)

4) the first child is a girl and the second child is a girl (gg)

But already given that one child is boy. So we have three possibilities of (bb)(bg)(gb).

n(E)= both are boys=BB=1

n(S)= 3

Required probability P = n(E)/n(S) = 1/3.

A) 3/91 | B) 2/73 |

C) 1/91 | D) 3/73 |

Explanation:

In a circle of n different persons, the total number of arrangements possible = (n - 1)!

Total number of arrangements = n(S) = (15 – 1)! = 14 !

Taking three persons as a unit, total persons = 13 (in 4 units)

Therefore no. of ways for these 13 persons to around the circular table = (13 - 1)! = 12!

In any given unit, 3 particular person can sit in 3!. Hence total number of ways that any three person can sit =

n(E) = 12! X 3!

Therefore P (E) = probability of three persons sitting together = n(E) / n(S) = 12! X 3 ! / 14!

= 12!x3x2 / 14x13x12! = 6/14x13 = 3/91