**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\(\left(\frac{49^9}{50^{10}}\right)\)