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Probability

Question
CBSEENMA12033780

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?

Solution

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.
                                              

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