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) 1/7 | B) 2/7 |

C) 1/2 | D) 3/2 |

Explanation:

A leap year has 52 weeks and two days

Total number of cases = 7

Number of favourable cases = 1

i.e., {Friday, Saturday}

Required Probability = 1/7

A) 16/19 | B) 1 |

C) 3/2 | D) 17/20 |

Explanation:

n(S) = 20

n(Even no) = 10 = n(E)

n(Prime no) = 8 = n(P)

P(E U P) = 10/20 + 8/20 - 1/20 = 17/20

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

C) 1/2 | D) 34/7 |

Explanation:

The number of exhaustive events = 100 C₁ = 100.

We have 25 primes from 1 to 100.

Number of favourable cases are 75.

Required probability = 75/50 = 3/2.

A) 6/7 | B) 19/21 |

C) 7/31 | D) 5/21 |

Explanation:

Number of ways of (selecting at least two couples among five people selected) = (⁵C₂ x ⁶C₁)

As remaining person can be any one among three couples left.

Required probability = (⁵C₂ x ⁶C₁)/¹⁰C₅

= (10 x 6)/252 = 5/21

A) 1 | B) 1/2 |

C) 0 | D) 3/5 |

Explanation:

The number of exhaustive outcomes is 36.

Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then = 18/36 = 1/2

P(E) = 1 - 1/2 = 1/2.