A) 17/112 | B) 55/221 |

C) 55/121 | D) 33/221 |

Explanation:

n(S) = ${}^{52}C_{2}$ = 1326

Let A = event of getting both red cards

and B = event of getting both queens

then $\left(A\cap B\right)$ = event of getting two red queens

n(A) = ${}^{26}C_{2}$ = 325, n(B) = ${}^{4}C_{2}$ = 6

$n(A\cap B)={}^{2}C_{2}=1$

$\therefore P\left(A\right)=\frac{325}{1326},P\left(B\right)=\frac{6}{1326}$

$P\left(A\cap B\right)=\frac{1}{1326}$

$\therefore $P ( both red or both queens) = $P\left(A\cup B\right)$

= $P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)=$$\frac{325}{1326}+\frac{1}{221}-\frac{1}{1326}$=$\frac{55}{221}$

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

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 =

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

C) 1/52 | D) 13/51 |

Explanation:

We know that:

When one card is drawn from a pack of 52 cards

The numbers of possible outcomes n(s) = 52

We know that there are 26 red cards in the pack of 52 cards

⇒ The numbers of favorable outcomes n(E) = 26

Probability of occurrence of an event: P(E)=Number of favorable outcomes/Numeber of possible outcomes=n(E)/n(S)

∴ required probability = 26/52 = 1/2.

A) (1/2)^11 | B) (9)(1/2) |

C) (11C2)(1/2)^9 | D) (1/2) |

Explanation:

Probability of occurrence of an event,

P(E) = Number of favorable outcomes/Numeber of possible outcomes = n(E)/n(S)

⇒ Probability of getting head in one coin = ½,

⇒ Probability of not getting head in one coin = 1- ½ = ½,

Hence,

All the 11 tosses are independent of each other.

∴ Required probability of getting only 2 times heads =${\left(\frac{1}{2}\right)}^{2}\times {\left(\frac{1}{2}\right)}^{9}={\left(\frac{1}{2}\right)}^{11}$