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

C) 2/5 | D) 4/15 |

Explanation:

Here Sample space S = { HH, HT, T1, T2, T3, T4, T5, T6 }

Let A be the event that the die shows a number greater than 4 and B be the event that the first throw of the coin results in a tail then,

A = { T5, T6 }

B = { T1, T2, T3, T4, T5, T6 }

Therefore, Required probability =

A) 3/7 | B) 7/11 |

C) 5/9 | D) 6/13 |

Explanation:

Here n(S) = 6 x 6 = 36

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

=> n(E)=20

Required Probability n(P) = n(E)/n(S) = 20/36 = 5/9.

A) A gain of Rs. 27 | B) A loss of Rs. 37 |

C) A loss of Rs. 27 | D) A gain of Rs. 37 |

Explanation:

As the win leads to multiplying the amount by 1.5 and loss leads to multiplying the amount by 0.5, we will multiply the initial amount by 1.5 thrice and by 0.5 thrice (in any order).

The overall resultant will remain same.

So final amount with the person will be (in all cases):

= 64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5)= Rs. 27

Hence the final result is:

64 − 27 = 37

A loss of **Rs.37**

A) 1/26 | B) 1/13 |

C) 2/13 | D) 1/52 |

Explanation:

Here in this pack of cards, n(S) = 52

Let E = event of getting a queen of the club **or** a king of the heart

Then, n(E) = 2

P(E) = **n(E)/n(S)** = 2/52 = 1/26

A) 14/33 | B) 14/55 |

C) 12/55 | D) 13/33 |

Explanation:

Total number of chairs = (3 + 5 + 4) = 12.

Let S be the sample space.

Then, n(s)= Number of ways of picking 2 chairs out of 12

= 12×11/2×1 = 66

Let n(E) = number of events of selecting 2 chairs for selecting no white chairs.

=> 8C2 = 8×7/2×1 = 28

Therefore required probability = 28/66 = 14/33.

A) 37/64 | B) 27/56 |

C) 11/13 | D) 9/8 |

Explanation:

Probability of not hitting the ball = 1- 1/4 =

Then, the probability that the batsman will hit the ball =