1,329 views
1 votes
1 votes
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 ____.

1 Answer

3 votes
3 votes

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

edited by