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Probability

Question
CBSEENMA12033740

Two persons throw a die alternatively till one of them gets a three and wins the game. Find their respective probability of winning.

Solution
Let E be the event of getting 3 in a throw of die.
   therefore space space straight P left parenthesis straight E right parenthesis space equals space 1 over 6 comma space space straight P left parenthesis straight E with bar on top right parenthesis space equals space 1 minus straight P left parenthesis straight E right parenthesis space equals space 1 minus 1 over 6 space equals 5 over 6

Any person who starts the game can win in the first throw, 3rd throw, 5th throw and so on.
therefore    probability of winning A
                               space equals space straight P left parenthesis straight E right parenthesis space plus space straight P left parenthesis straight E with bar on top space straight E with bar on top space straight E right parenthesis space plus space straight P left parenthesis straight E with bar on top space straight E with bar on top space straight E with bar on top space straight E with bar on top straight E right parenthesis plus...
equals space 1 over 6 cross times 1 over 6 cross times 5 over 6 cross times 5 over 6 plus 1 over 6 cross times 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 5 over 6 plus...
equals space 1 over 6 open square brackets 1 plus open parentheses 5 over 6 close parentheses squared plus open parentheses 5 over 6 close parentheses to the power of 4 plus.... close square brackets
equals space 1 over 6 open square brackets fraction numerator 1 over denominator 1 minus begin display style 25 over 36 end style end fraction close square brackets space equals space 1 over 6 cross times 36 over 11 space equals space 6 over 11
Since either of two person wins
    therefore         probability of winning B = 1 minus 6 over 11 space equals 5 over 11
   therefore          required probabilities are 6 over 11 comma space 5 over 11.

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