# Probability Questions

A) 1/2 | B) 7/15 |

C) 8/15 | D) 1/9 |

Explanation:

Let S be the sample space

Then n(S) = no of ways of drawing 2 balls out of (6+4) = = =45

Let E = event of getting both balls of same colour

Then,n(E) = no of ways (2 balls out of six) or (2 balls out of 4)

= = = 15+6 = 21

Therefore, P(E) = n(E)/n(S) = 21/45 = 7/15

A) 52/221 | B) 55/190 |

C) 55/221 | D) 19/221 |

Explanation:

We have n(s) = = = 1326.

Let A = event of getting both black cards

B = event of getting both queens

A∩B = event of getting queen of black cards

n(A) = = = 325, n(B)= = = 6 and n(A∩B) = = 1

P(A) = n(A)/n(S) = 325/1326;

P(B) = n(B)/n(S) = 6/1326 and

P(A∩B) = n(A∩B)/n(S) = 1/1326

P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

A) 2/7 | B) 1/7 |

C) 3/4 | D) 4/5 |

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) 2/7 | B) 5/7 |

C) 1/5 | D) 1/2 |

Explanation:

Total number of outcomes possible, n(S) = 10 + 25 = 35

Total number of prizes, n(E) = 10