A box contains 2 washers, 3 nuts and 4 bolts. Items are drawn from the box at random one at a time without replacement. The probability of drawing 2 washers first followed by 3 nuts and subsequently the 4 bolts is

2/315

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1/630

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1/1260

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1/2520

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Solution

The correct option is C 1/1260
Given,

$Box \begin{matrix} 2 washers 3 nuts 4 bolts \end{matrix}$
The required probability
$= (\frac{2}{9} \times \frac{1}{8}) \times (\frac{3}{7} \times \frac{2}{6} \times \frac{1}{5}) \times (\frac{4}{4} \times \frac{3}{3} \times \frac{2}{2} \times \frac{1}{1})$

$= \frac{1}{36} \times \frac{1}{35} = \frac{1}{1260}$

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A box contains 2 washers, 3 nuts and 4 bolts. Items are drawn from the box at random one at a time without replacement. The probability of drawing 2 washers first followed by 3 nuts and subsequently the 4 bolts is

The correct option is C 1/1260Given, Box2 washers3 nuts4 bolts The required probability =(29×18)×(37×26×15)×(44×33×22×11) =136×135=11260