A) 1/3 | B) 1/9 |

C) 8/9 | D) 9/10 |

Explanation:

S = { (1, 1), (1, 2), (1, 3), (1, 4),(1, 5), (1, 6), (2, 1), (2, 2),.........(6, 5), (6, 6) }

=> n(S) = 6 x 6 = 36

E = {(6, 3), (5, 4), (4, 5), (3, 6) }

=> n(E) = 4

Therefore, P(E) = 4/36 = 1/9

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

A) 1 | B) 1/5 |

C) 5 | D) 4/5 |

Explanation:

Number of applicants = 5

On a day, only 1 leave is approved.

Now favourable events = 1 of 5 applicants is approved

Probability that Laxmi priya's leave is granted = 1/5.

A) 6/19 | B) 3/10 |

C) 7/10 | D) 6/17 |

Explanation:

Here, S = {1, 2, 3, 4, ...., 19, 20}=> n(s) = 20

Let E = event of getting a multiple of 4 or 15

=multiples od 4 are {4, 8, 12, 16, 20}

And multiples of 15 means multiples of 3 and 5

= {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}.

= the common multiple is only (15).

=> E = n(E)= 6

Required Probability = P(E) = n(E)/n(S) = 6/20 = 3/10.