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# 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} Can you answer this question? ## 1 Answer 0 votes 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