# A factory produces bulbs,The probability that any one bulb is defective is $\Large \frac{1}{50}$ and they are packed in boxes of 10.From a single box,find the probability that\begin{array}{1 1}(i)\;none\;of\;the\;bulbs\;is \;defective.\$$ii)\;exactly\;two\;bulbs\;are\;defective\\(iii)\;more\;than\;8\;bulbs\;work\;properly\end{array} ## 1 Answer Need homework help? Click here. Toolbox: • In a factory probability of defective bulbs is • P(defective)=\(\large\frac{1}{50}$$
• p=$$\large\frac{1}{50}$$ $$q=1-P$$ =$$\large\frac{49}{50}$$
• A box contains 10 bulbs
• n=10
• P(x=r)=$$^n$$c$$_r$$ p$$^r$$q$$^{n-r}$$
P(none defective)=$$p(x=0)$$
=$$^{10}$$c$$_0$$($$\frac{1}{50}$$)$$^0$$($$\frac{49}{50}$$)$$^{10}$$
=($$\frac{49}{50}$$)$$^{10}$$
p(exactly 2defective)=P(x=2)
=$$^{10}$$c$$_2$$($$\frac{1}{50}$$)$$^2$$($$\frac{49}{50}$$)$$^8$$
=45$$\left(\frac{49^8}{50^{10}}\right)$$
P(more then 8 working property)
=P(less than 2 not working)
=P(less than 2 defeective)
=P(x=0)+P(x=1)
=$$^{10}$$c$$_0$$($$\frac{1}{50}$$)$$^0$$($$\frac{49}{50}$$)$$^{10}$$+$$^{10}$$c$$_1$$($$\frac{1}{50}$$)$$^1$$($$\frac{49}{50}$$)$$^9$$
=($$\frac{49}{50}$$)$$^9$$+10x$$\left(\frac{49^9}{50^{10}}\right)$$
=$$\left(\frac{49^9}{50^{10}}\right)$$[49+10]
=59$$\left(\frac{49^9}{50^{10}}\right)$$

edited Mar 15, 2013