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Probability

Question
CBSEENMA12033765

There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

Solution

Let E1, E2, E3, E be the events
E1 : ‘coin chosen is two headed’,
E2 : ‘coin chosen is biased’,
E: ‘coin chosen is unbiased’,
E : ‘tossed coin shows up a head’
                 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 1 third.
                straight P left parenthesis straight E space vertical line thin space straight E subscript 1 right parenthesis space equals space 1 comma space space space straight P left parenthesis straight E space vertical line thin space straight E subscript 2 right parenthesis space equals space 75 over 100 equals space 3 over 4 comma space space space straight P left parenthesis straight E space vertical line space straight E subscript 3 right parenthesis space equals 1 half
Required probability  = 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 thin space vertical line thin space 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 space vertical line thin space straight E subscript 1 right parenthesis thin space plus space straight P left parenthesis straight E subscript 2 right parenthesis thin space straight P left parenthesis straight E space vertical line thin space 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 space vertical line thin space straight E subscript 3 right parenthesis end fraction left parenthesis By space Baye apostrophe straight s space Theorem right parenthesis
equals space fraction numerator begin display style 1 third end style cross times 1 over denominator begin display style 1 third end style cross times 1 cross times begin display style 1 third end style cross times begin display style 3 over 4 end style plus begin display style 1 third end style cross times begin display style 1 half end style end fraction space equals fraction numerator 1 over denominator 1 plus begin display style 3 over 4 end style plus begin display style 1 half end style end fraction equals space fraction numerator 1 over denominator begin display style fraction numerator 4 plus 3 plus 2 over denominator 4 end fraction end style end fraction space equals 4 over 9 

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