A) 5/7 | B) 1/5 |

C) 2/7 | D) 2/35 |

Explanation:

P( only one of them will be selected)

= p[(E and not F) or (F and not E)]

=

=

=

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 =

A) 1/7 | B) 8! |

C) 7! | D) 1/14 |

Explanation:

The 8 letters can be written in 8! ways.

n(S) = 8!

Let E be the event that the letters b,c,d,e always come together when the first 8 alphabets are written down.

Now the letters a, (bcde), f, g and h can be arranged in 5! ways.

The letters b,c,d and e can be arranged themselves in 4! ways.

n(E) = 5! x 4!

Now, the required P(E) = n(E)/n(S) = 5! x 4!/8! = 1/14

Hence the answer is 1/14.