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

C) 2/5 | D) 4/15 |

Explanation:

Here Sample space S = { HH, HT, T1, T2, T3, T4, T5, T6 }

Let A be the event that the die shows a number greater than 4 and B be the event that the first throw of the coin results in a tail then,

A = { T5, T6 }

B = { T1, T2, T3, T4, T5, T6 }

Therefore, Required probability =

A) 35/96 | B) 19/90 |

C) 19/96 | D) None of these |

Explanation:

Here N and C are not common and same letters can be A, I, S, T. Therefore

Probability of choosing A = = 1/45

Probability of choosing I = = 1/45

Probability of choosing S = = 1/10

Probability of choosing T = = 1/15

Hence, Required probability =

A) 8/39 | B) 15/39 |

C) 12/13 | D) None of these |

Explanation:

P( selecting atleast one couple) = 1 - P(selecting none of the couples for the prize)

=

A) 23/42 | B) 19/42 |

C) 7/32 | D) 16/39 |

Explanation:

A red ball can be drawn in two mutually exclusive ways

(i) Selecting bag I and then drawing a red ball from it.

(ii) Selecting bag II and then drawing a red ball from it.

Let E1, E2 and A denote the events defined as follows:

E1 = selecting bag I,

E2 = selecting bag II

A = drawing a red ball

Since one of the two bags is selected randomly, therefore

P(E1) = 1/2 and P(E2) = 1/2

Now, = Probability of drawing a red ball when the first bag has been selected = 4/7

= Probability of drawing a red ball when the second bag has been selected = 2/6

Using the law of total probability, we have

P(red ball) = P(A) =

=

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) 0.21 | B) 0.3 |

C) 0.7 | D) None of these |

Explanation:

P(atleast B) = P( B or A) = P(B) + P(A) = (0.3) + (0.4) = 0.7