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) 4/7 | B) 2/3 |

C) 1/2 | D) 5/6 |

Explanation:

Total coins 30

In that,

1 rupee coins 20

50 paise coins 10

Probability of total 1 rupee coins = 20C11

Probability that 11 coins are picked = 30C11

Required probability of a coin now picked from the box is 1 rupee **= 20C11/30C11 = 2/3.**

A) 15 | B) 18 |

C) 20 | D) 24 |

Explanation:

We know that, Total probability = 1

Given probability of black stones = 1/4

=> Probability of blue and white stones = 1 - 1/4 = 3/4

But, given blue + white stones = 9 + 6 = 15

Hence,

3/4 ----- 15

1 ----- ?

=> 15 x 4/3 = 20.

Hence, total number of stones in the box = **20.**

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) 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**