The probability of student A passing an examination is 2/9 and of student B passing is 5/9. Assuming the two events : 'A passes', 'B passes' as independent, find the probability of : (i) only A passing the examination (ii) only one of them passing the examination.

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Solution

$P (A passing examination) = \frac{2}{9} P (B passing examination) = \frac{5}{9} (i) P (only A passing examination) = P (A passes) P (B fails) = \frac{2}{9} (1 - \frac{5}{9}) = \frac{2}{9} \times \frac{4}{9} = \frac{8}{81} (ii) P (only one of them passing examination) = P (A passes and B fails) + P (B passes and A fails) = \frac{2}{9} \times (1 - \frac{5}{9}) + \frac{5}{9} \times (1 - \frac{2}{9}) = \frac{8}{81} + \frac{35}{81} = \frac{43}{81}$

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The probability of student A passing an examination is 2/9 and of student B passing is 5/9. Assuming the two events : 'A passes', 'B passes' as independent, find the probability of : (i) only A passing the examination (ii) only one of them passing the examination.

PA passing examination=29PB passing examination=59i Ponly A passing examination=PA passes PB fails=291-59=29×49=881ii Ponly one of them passing examination=PA passes and B fails+PB passes and A fails=29×1-59+59×1-29=881+3581=4381