A) 4/3 | B) 2/3 |

C) 3/2 | D) 3/4 |

Explanation:

Let S be the sample space.

Here n(S)= ${\mathbf{2}}^{\mathbf{3}}$ = 8

Let E be the event of getting atmost two heads. Then,

n(E) = {(H,T,T), (T,H,T), (T,T,H), (H,H,T), (T,H,H), (H,T,H)}

Required probability = n(E)/n(S) = 6/8 = 3/4.

A) 0 | B) -1 |

C) 0.1 | D) 1 |

Explanation:

The probability of an impossible event is 0.

The event is known ahead of time to be not possible, therefore by definition in mathematics, the probability is defined to be 0 which means it can never happen.

The probability of a certain event is 1.

A) 2/9 | B) 5/9 |

C) 4/9 | D) 0 |

Explanation:

Number of white marbles = 4

Number of Black marbles = 5

Total number of marbles = 9

Number of ways, two marbles picked randomly = 9C2

Now, the required probability of picked marbles are to be of same color = 4C2/9C2 + 5C2/9C2

= 1/6 + 5/18

= 4/9.

A) 2/3 | B) 1/8 |

C) 3/8 | D) 3/4 |

Explanation:

Given number of balls = 3 + 5 + 7 = 15

One ball is drawn randomly = 15C1

probability that it is either pink or red = $\frac{\mathbf{7}{\mathbf{C}}_{\mathbf{1}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{3}{\mathbf{C}}_{\mathbf{1}}}{\mathbf{15}{\mathbf{C}}_{\mathbf{1}}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{7}\mathbf{}\mathbf{+}\mathbf{}\mathbf{3}}{\mathbf{15}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{10}}{\mathbf{15}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{2}}{\mathbf{3}}$

A) 1/4 | B) 1/6 |

C) 1/8 | D) 4 |

Explanation:

Required probability is given by P(E) = $\frac{\mathbf{n}\mathbf{\left(}\mathbf{E}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{C}}_{\mathbf{2}}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{6}}$

A) 11/379 | B) 21/628 |

C) 24/625 | D) 26/247 |

Explanation:

Total no of ways = (14 – 1)! = 13!

Number of favorable ways = (12 – 1)! = 11!

So, required probability = $\left(\frac{\left(\mathbf{11}\mathbf{!}\mathbf{\times}\mathbf{3}\mathbf{!}\right)}{\mathbf{13}\mathbf{!}}\right)$ = $\frac{39916800\times 6}{6227020800}$ = $\frac{\mathbf{24}}{\mathbf{625}}$

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