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

C) 3/4 | D) 7/12 |

Explanation:

Let A, B, C be the respective events of solving the problem and be the respective events of not solving the problem. Then A, B, C are independent events

are independent events

Now, P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4

P( none solves the problem) = P(not A) and (not B) and (not C)

=

= =

=

Hence, P(the problem will be solved) = 1 - P(none solves the problem)

=

A) 1 | B) 1/5 |

C) 5 | D) 4/5 |

Explanation:

Number of applicants = 5

On a day, only 1 leave is approved.

Now favourable events = 1 of 5 applicants is approved

Probability that Laxmi priya's leave is granted = 1/5.

A) 6/19 | B) 3/10 |

C) 7/10 | D) 6/17 |

Explanation:

Here, S = {1, 2, 3, 4, ...., 19, 20}=> n(s) = 20

Let E = event of getting a multiple of 4 or 15

=multiples od 4 are {4, 8, 12, 16, 20}

And multiples of 15 means multiples of 3 and 5

= {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}.

= the common multiple is only (15).

=> E = n(E)= 6

Required Probability = P(E) = n(E)/n(S) = 6/20 = 3/10.

A) 8/15 | B) 7/15 |

C) 1/15 | D) 4/15 |

Explanation:

Given total students in the class = 60

Students who are taking Economics = 24 and

Students who are taking Calculus = 32

Students who are taking both subjects = 60-(24 + 32) = 60 - 56 = 4

Students who are taking calculus only = 32 - 4 = 28

probability that a randomly chosen student from this group is taking only the Calculus class = 28/60 = 7/15.

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

C) 3/2 | D) 3/4 |

Explanation:

Let S be the sample space.

Here n(S)= = 8

Let E be the event of getting atmost two heads. Then,

n(E) = {(H,T,T), (T,H,T), (T,T,H), (H,H,T), (T,H,H), (H,T,H)}

Required probability = n(E)/n(S) = 6/8 = 3/4.

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

C) 7/8 | D) 4/5 |

Explanation:

Multiples of 3 below 20 are 3, 6, 9, 12, 15, 18

Multiples of 5 below 20 are 5, 10, 15, 20

Required number of possibilities = 10

Total number of possibilities = 20

Required probability = 10/20 = 1/2.