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Solution :

P(getting a defective bulb) `=(5)/(100) =(1)/(20)` and <br> P(getting a nondefective bulb) `=(1-(1)/(20)) =(19)/(20)` <br> Then `p =(1)/(20) " and " q=(19)/(20)` <br> Let X denote the number of defective bulbs <br> `P(X=r)=.^(n)C_(r ).p^(r).q^((n-r)) =.^(10)C_(r ).((1)/(20))^( r) .((19)/(20))^((10-r))` <br> P(getting not more than 2 defective bulbs) <br> `=P(X=0 " or " X=1 " or " X=2)` <br> `=P(X =0) +P(X=1)+P(X=2)` <br> `=.^(10)C_(0).((1)/(20))^(0).((19)/(20))^(10) +.^(10)C_(1).((1)/(20))^(1).((19)/(20))^(9)+.^(10)C_(2).((1)/(20))^(2).((19)/(20))^(8)` <br> `=((19)/(20))^(10)=(1)/(2).((19)/(20))^(9)+(9)/(80).((19)/(20))^(8)=((19)/(20))^(8) .((149)/(100))` <br> Let `A =((19)/(20))^(8) .((149)/(100)) .` Then <br> log A =8(log 19-log 20) +log 149 -log 100 <br> `=8(1.2788 -1.3010 )+ 2.1732 -2 ` <br> `=-0.0044 =1.9956` <br> `:. A =` antilog `(bar(1).9956)=0.99.` <br> Hence the required probability `=(99)/(100)`