A) 1/2 | B) 3/4 |

C) 1/9 | D) 2/9 |

Explanation:

In two throws of a die, n(S) = (6 x 6) = 36.

Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.

P(E) =n(E)/n(S)=4/36=1/9.

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) 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.