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Neeraj Sharma 8 years, 9 months ago
{tex}let\,P\left( B \right) = x{/tex}
{tex}then\,P\left( A \right) = 6{\left[ {P\left( B \right)} \right]^2} = 6{x^2}{/tex}
{tex}Since\,A\,and\,B\,are\,mutually\,exclusive\,and\,exhaustive\,events{/tex}
{tex}therefore{/tex}
{tex}A \cup B = S{/tex}
{tex}P\left( {A \cup B} \right) = P\left( S \right){/tex}
{tex}P\left( {A \cup B} \right) = 1{/tex}
{tex}P\left( A \right) + P\left( B \right) = 1{/tex}
{tex}6{x^2} + x = 1{/tex}
{tex}6{x^2} + x - 1 = 0{/tex}
{tex}\left( {2x + 3} \right)\left( {3x - 1} \right) = 0{/tex}
{tex}x = - {2 \over 3},{1 \over 3}{/tex}
{tex}hence{/tex}
{tex}P\left( B \right) = {1 \over 3}{/tex}
{tex}P\left( A \right) = 6 \times {\left( {{1 \over 3}} \right)^2} = {2 \over 3}{/tex}
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