A) 35/96 | B) 19/90 |

C) 19/96 | D) None of these |

Explanation:

Here N and C are not common and same letters can be A, I, S, T. Therefore

Probability of choosing A = = 1/45

Probability of choosing I = = 1/45

Probability of choosing S = = 1/10

Probability of choosing T = = 1/15

Hence, Required probability =

A) 1/7 | B) 8! |

C) 7! | D) 1/14 |

Explanation:

The 8 letters can be written in 8! ways.

n(S) = 8!

Let E be the event that the letters b,c,d,e always come together when the first 8 alphabets are written down.

Now the letters a, (bcde), f, g and h can be arranged in 5! ways.

The letters b,c,d and e can be arranged themselves in 4! ways.

n(E) = 5! x 4!

Now, the required P(E) = n(E)/n(S) = 5! x 4!/8! = 1/14

Hence the answer is 1/14.

A) 1/36 | B) 5/36 |

C) 1/12 | D) 1/9 |

Explanation:

The two events mentioned are independent.

The first roll of the die is independent of the second roll. Therefore the probabilities can be directly multiplied.

P(getting first 4) = 1/6

P(no second 6) = 5/6

Therefore P(getting first 4 and no second 6) = 1/6 x 5/6 = 5/36

A) TRUE | B) FALSE |

Explanation:

These two events cannot be disjoint because P(K) + P(L) > 1.

P(AꓴB) = P(A) + P(B) - P(AꓵB).

An event is disjoint if P(A ꓵ B) = 0. If K and L are disjoint P(K ꓴ L) = 0.8 + 0.6 = 1.4

And Since probability cannot be greater than 1, these two mentioned events cannot be disjoint.

A) 1/7 | B) 2/7 |

C) 1/2 | D) 3/2 |

Explanation:

A leap year has 52 weeks and two days

Total number of cases = 7

Number of favourable cases = 1

i.e., {Friday, Saturday}

Required Probability = 1/7

A) 16/19 | B) 1 |

C) 3/2 | D) 17/20 |

Explanation:

n(S) = 20

n(Even no) = 10 = n(E)

n(Prime no) = 8 = n(P)

P(E U P) = 10/20 + 8/20 - 1/20 = 17/20