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A packet of 10 bulbs is known to include 2 bulbs that are defective. If 4 bulbs are randomly chosen and tested, the probability of finding among them not more than 1 defective bulb is ____.
asked in Combinatory by (257 points)   | 119 views
No of ways To choose 4 nondefective bulbs: 8C4

No of ways To choose only 1 defective bulbs: 8C3x2C1

Probability of atmost 1 defective bulbs: [8C4x(8C3x2C1)] / 10C4
your answer comes out to be 0.86 but the given answer ranges from 0.80 to 0.82

1 Answer

+1 vote

P(X==0) + P(X==1)

= $(\frac{8}{10})^{4}+\binom{4}{1}*(\frac{2}{10})*(\frac{8}{10})^{4}$

= 0.8192

------------------------------------------------

Above answer is wrong since it considers the trails with replacement. By default, it should be without replacement(unless specified)

So, answer = $\frac{\binom{8}{4}}{\binom{10}{4}} + \frac{\binom{8}{3}*\binom{2}{1}}{\binom{10}{4}}$

                   = 0.86

answered by Veteran (15.4k points)  
edited by

sushant..whats wrong with this

8C4/10C4 (when no one is detective) + 2C1 + 8C3/ 10C4

 

When you do choose operation i.e. nCx, you are choosing one-by-one without replacement.

The above problem is clear indication of Bernoulli trial (where the trials are finite). Also, in bernoulli trial, you consider with replacement.

@Akriti. Let me think. Now, I am confused :)
yea suree..i always get confused in such questions..btw this is the similar question http://gateoverflow.in/88215/bulbs-out-sample-bulbs-manufactured-company-are-defective
but according to the language of the quesiton,i thinkit should be without replacement as 4 are picked randomly and default case is with replacment only

@Akriti. Yes. It should be without replacement.

So, 0.86 is the right answer.

WIll edit.

But, MadeEasy has given the solution to be 0.82 using bernoulli's trials
No ,it should be wrong.

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