Suppose we have four boxes A, B, C and D containing coloured mar

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsShort Answer Type

931. A man is known to speak the truth 5 out of 6 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.
74 Views

 Multiple Choice QuestionsLong Answer Type

932. Let there be three urns containing 1 white, 2 black, 3 red balls ; 2 white, 1 black, 1 red ball ; 4 white, 5 black, 3 red balls. One urn is chosen at random and two balls are drawn. These happen to be white and a red. What is the probability that they come from urn number 3.
96 Views

933. A pack of playing cards was found to contain only 51 cards. If the first 13 cards which are examined are all red, what is the probability that the missing card is black.
149 Views

934.

Coloured balls are distributed in four boxes as shown in the following table:

Box

 

Colour

 
 

Black

White

Red

Blue

I

3

4

5

6

II

2

2

2

2

III

1

2

3

1

IV

4

3

1

5

 

A box is selected at random and then a ball is randomly drawn from the selected box. The colour of the ball is black, what is the probability that ball drawn is from the box III?
85 Views

Advertisement
Advertisement

935.

Suppose we have four boxes A, B, C and D containing coloured marbles as given below:

Box

Marble Colour

 

Red

White

Black

A

1

6

3

B

6

2

2

C

8

1

1

D

0

6

4

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red. what is the probability that it was drawn from box A? box B? box C?


Let E, E1, E2, E3 and E4 be the events as defined below:
E : ‘ball drawn is red’, E1 : ‘box A is selected’, E2: ‘box B is selected’,
E: ‘box C is selected ‘ and E4 : ‘box D is selected’,
therefore space space space space straight P left parenthesis straight E subscript 1 right parenthesis space equals space straight P left parenthesis straight E subscript 2 right parenthesis space equals space straight P left parenthesis straight E subscript 3 right parenthesis space equals space straight P left parenthesis straight E subscript 4 right parenthesis space equals space 1 fourth
and  straight P left parenthesis straight E space vertical line thin space straight E subscript 1 right parenthesis space equals space fraction numerator 1 over denominator 1 plus 6 plus 3 end fraction space equals space 1 over 10

           straight P left parenthesis straight E thin space vertical line thin space straight E subscript 2 right parenthesis space equals space fraction numerator 6 over denominator 6 plus 2 plus 2 end fraction equals space 6 over 10
straight P left parenthesis straight E thin space vertical line thin space straight E subscript 3 right parenthesis space equals space fraction numerator 8 over denominator 8 plus 1 plus 1 end fraction equals space 8 over 10

and    straight P left parenthesis straight E space vertical line space straight E subscript 4 right parenthesis space equals space fraction numerator 0 over denominator 0 plus 6 plus 4 end fraction space equals 0

P(ball is drawn from box A) = straight P left parenthesis straight E subscript 1 space vertical line thin space straight E right parenthesis
                   equals space fraction numerator straight P left parenthesis straight E subscript 1 right parenthesis thin space straight P left parenthesis straight E vertical line straight E subscript 1 right parenthesis over denominator straight P left parenthesis straight E subscript 1 right parenthesis thin space straight P left parenthesis straight E vertical line straight E subscript 1 right parenthesis space plus space straight P left parenthesis straight E subscript 2 right parenthesis thin space straight P left parenthesis straight E vertical line straight E subscript 2 right parenthesis space plus space straight P left parenthesis straight E subscript 3 right parenthesis thin space straight P left parenthesis straight E vertical line straight E subscript 3 right parenthesis space plus space straight P left parenthesis straight E subscript 4 right parenthesis thin space straight P left parenthesis straight E vertical line straight E subscript 4 right parenthesis end fraction

                                                                                open square brackets because space of space Baye apostrophe straight s space Theorem close square brackets
                equals space fraction numerator begin display style 1 fourth end style cross times begin display style 1 over 10 end style over denominator begin display style 1 fourth end style cross times begin display style 1 over 10 end style plus begin display style 1 fourth end style cross times begin display style 6 over 10 end style plus begin display style 1 fourth end style cross times begin display style 8 over 10 end style plus begin display style 1 fourth end style cross times 0 end fraction space equals fraction numerator 1 over denominator 1 plus 6 plus 8 end fraction space equals 1 over 15

P(ball is drawn from box C) = straight P left parenthesis straight E subscript 2 vertical line straight E right parenthesis

equals space fraction numerator straight P left parenthesis straight E subscript 2 right parenthesis thin space straight P left parenthesis straight E vertical line straight E subscript 2 right parenthesis over denominator straight P left parenthesis straight E subscript 1 right parenthesis thin space straight P left parenthesis straight E vertical line straight E subscript 1 right parenthesis space plus space straight P left parenthesis straight E subscript 2 right parenthesis thin space straight P left parenthesis straight E vertical line straight E subscript 2 right parenthesis space plus space straight P left parenthesis straight E subscript 3 right parenthesis thin space straight P left parenthesis straight E vertical line straight E subscript 3 right parenthesis space plus space straight P left parenthesis straight E subscript 4 right parenthesis thin space straight P left parenthesis straight E vertical line straight E subscript 4 right parenthesis end fraction

                                                                       open square brackets because space space of space Baye apostrophe straight s space Theorem close square brackets
equals space fraction numerator begin display style 1 fourth end style cross times begin display style 6 over 10 end style over denominator begin display style 1 fourth end style cross times begin display style 1 over 10 end style plus begin display style 1 fourth end style cross times begin display style 6 over 10 end style plus begin display style 1 fourth end style cross times begin display style 8 over 10 end style plus begin display style 1 fourth end style cross times 0 end fraction space equals fraction numerator 6 over denominator 1 plus 6 plus 8 end fraction space equals 6 over 15

and P(ball is drawn from box A) =  straight P left parenthesis straight E subscript 3 vertical line straight E right parenthesis
 equals space fraction numerator straight P left parenthesis straight E subscript 3 right parenthesis thin space straight P left parenthesis straight E vertical line straight E subscript 3 right parenthesis over denominator straight P left parenthesis straight E subscript 1 right parenthesis thin space straight P left parenthesis straight E vertical line straight E subscript 1 right parenthesis space plus space straight P left parenthesis straight E subscript 2 right parenthesis thin space straight P left parenthesis straight E vertical line straight E subscript 2 right parenthesis space plus space straight P left parenthesis straight E subscript 3 right parenthesis thin space straight P left parenthesis straight E vertical line straight E subscript 3 right parenthesis space plus space straight P left parenthesis straight E subscript 4 right parenthesis thin space straight P left parenthesis straight E vertical line straight E subscript 4 thin space right parenthesis end fraction

                                                             open square brackets because space space of space Baye apostrophe straight s space Theorem close square brackets
equals space fraction numerator begin display style 1 fourth end style cross times begin display style 8 over 10 end style over denominator begin display style 1 fourth end style cross times begin display style 1 over 10 end style plus begin display style 1 fourth end style cross times begin display style 6 over 10 end style plus begin display style 1 fourth end style cross times begin display style 8 over 10 end style plus begin display style 1 fourth end style cross times 0 end fraction space equals fraction numerator 8 over denominator 1 plus 6 plus 8 end fraction space equals 8 over 15.

                                              

84 Views

Advertisement

 Multiple Choice QuestionsMultiple Choice Questions

936. Probability that A speaks truth is 4 over 5. A coin is tossed. A reports that a head appears. The probability that actually there was head is 
  • 4 over 5
  • 1 half
  • 1 fifth
  • 1 fifth
78 Views

 Multiple Choice QuestionsShort Answer Type

937. A person plays a game of tossing a coin thrice. For each head, he is given Rs. 2 by the organiser of the game and for each tail, he has to give Rs. 1.50 to the organiser. Let X denote the amount gained or lost by the person. Show that X is a random variable and exhibit it as a function on the sample space of the experiment.
91 Views

938. An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X ? Is X a random variable?
115 Views

Advertisement
939. Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X?
87 Views

940. A bag contains 2 white and 1 red balls. One ball is drawn at random and then put back in the box after noting its colour. The process is repeated again. If X denotes the number of red balls recorded in the two draws, describe X.
102 Views

Advertisement