A) 64 | B) 84 |

C) 56 | D) 20 |

Explanation:

Total number of possible ways = 9C3 = 84 ways

Required atleast one girl in the group of three = total possible ways - ways in which none is girl

None of the members in the group is girl = 6C3 = 20

Therefore, number of ways that at least one member is a girl in the group of three

= 84 - 20

**= 64 ways.**

A) 1 | B) 21/26 |

C) 5/26 | D) 5 |

Explanation:

We know that,

Total number of balls n(S) = 26

Number of vowels n(E) = 5

Hence, required probability = n(E)/n(S) = **5/26.**

A) 110/129 | B) 5/6 |

C) 121/126 | D) 3/7 |

Explanation:

Total number of possible ways =

$\mathbf{9}{\mathbf{C}}_{\mathbf{5}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{9}\mathbf{}\mathbf{x}\mathbf{}\mathbf{8}\mathbf{}\mathbf{x}\mathbf{}\mathbf{7}\mathbf{}\mathbf{x}\mathbf{}\mathbf{6}\mathbf{}\mathbf{x}\mathbf{}\mathbf{5}}{\mathbf{5}\mathbf{}\mathbf{x}\mathbf{}\mathbf{4}\mathbf{}\mathbf{x}\mathbf{}\mathbf{3}\mathbf{}\mathbf{x}\mathbf{}\mathbf{2}\mathbf{}\mathbf{x}\mathbf{}\mathbf{1}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{126}$

Number of favorable cases =

$\mathbf{4}{\mathbf{C}}_{\mathbf{2}}\mathbf{}\mathbf{x}\mathbf{}\mathbf{5}{\mathbf{C}}_{\mathbf{3}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{4}{\mathbf{C}}_{\mathbf{3}}\mathbf{}\mathbf{x}\mathbf{}\mathbf{5}{\mathbf{C}}_{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{4}{\mathbf{C}}_{\mathbf{4}}\mathbf{}\mathbf{x}\mathbf{}\mathbf{5}{\mathbf{C}}_{\mathbf{1}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{4\mathrm{x}3}{2\mathrm{x}1}\mathrm{x}\frac{5\mathrm{x}4\mathrm{x}3}{3\mathrm{x}2\mathrm{x}1}+\frac{4\mathrm{x}3\mathrm{x}2}{3\mathrm{x}2\mathrm{x}1}\mathrm{x}\frac{5\mathrm{x}4}{2\mathrm{x}1}+1\mathrm{x}5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=6\mathrm{x}10+4\mathrm{x}10+5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=60+40+5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{}\mathbf{105}$

Therefore, required probability = **105/126 = 5/6**

A) 7 | B) 1/7 |

C) 3/7 | D) 0 |

Explanation:

Given total number of smileys = 5 + 6 + 3 = 14

Now, required probability =

$\frac{\mathbf{5}{\mathbf{C}}_{\mathbf{2}}}{\mathbf{14}{\mathbf{C}}_{\mathbf{2}}}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{3}{\mathbf{C}}_{\mathbf{2}}}{\mathbf{14}{\mathbf{C}}_{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{10}{91}+\frac{3}{91}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{13}{91}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{7}}$

A) 35/36 | B) 17/36 |

C) 15/36 | D) 1/36 |

Explanation:

When two dice are thrown simultaneously, the probability is **n(S)** = 6x6 = 36

Required, the sum of the two numbers that turn up is less than 12

That can be done as **n(E)**

=** {** (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)

(6,1), (6,2), (6,3), (6,4), (6,5)** }**

**= 35**

Hence, required probability =** n(E)/n(S) = 35/36.**

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.