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A box contain 2 washers, 3 nuts and 4 bolts.

items are drawn from the box at random one at a time w/o replacement.

the probability of drawing 2 washers first followed by 3 nuts and subsequently the 4 bolt is ???

$\frac{2}{9}*\frac{1}{8}*\frac{3}{7}*\frac{2}{6}*\frac{1}{5}*\frac{4}{4}*\frac{3}{3}*\frac{2}{2}*\frac{1}{1}$
can u explain????
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yes C)

$\frac{2!.3!.4!}{9!}=\frac{1}{1260}$

The given question deals with the case in which the sequence of events is mentioned and the items are drawn without replacement means the population is going to change every time and hence the probability of subsequent event.Let me make this point clear by taking the question :

First , a washer is taken out.So,

P(1st washer is taken out) = 1/(2+3+4) = 2/9

Since the term "without replacement" is used so now population(sample space) will now decrease to 8.Had these been "with replacement" then the cardinality of sample space will remain same i.e. 9 in this case.

So, P(2nd washer is taken out) = 1/(1+3+4) = 1/8

Similarly now P(1st nut is taken out) = 3/(3+4) = 3/7 [Since initially we have 3 nuts which is favorable case]

P(2nd nut is taken out)  =  2/6

P(3rd nut)      =     1/5

Now P(1st bolt)  =  4/4   [Since 4 bolts are present initially]

P(2nd bolt) =  3/3

P(3rd bolt)  = 2/2

P(4th bolt)  = 1/1

So by multiplication principle [Since all above mentioned events have to happen subsequently] , we have :

P(2 washers followed by 3 nuts followed by 4 bolts and without replacement]

=   2/9 * 1/8 * 3/7 * 2/6 * 1/5 * 4/4 * 3/3 * 2/2 * 1/1

=   (23 * 32 * 4) / 9!

here it is  = 2/9 * 1/8 * 3/7 * 2/6 * 1/5 * 4/4 * 3/3 * 2/2 * 1/1

i got it thks for correction

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