Question Number 55737 by byaw last updated on 03/Mar/19
Commented by mr W last updated on 03/Mar/19
$$\left({i}\right) \\ $$$${p}=\left(\mathrm{0}.\mathrm{3}\right)^{\mathrm{4}} =\mathrm{0}.\mathrm{0081} \\ $$$$\left({ii}\right) \\ $$$${p}=\left(\mathrm{0}.\mathrm{7}\right)^{\mathrm{4}} =\mathrm{0}.\mathrm{2401} \\ $$
Commented by byaw last updated on 03/Mar/19
$${Thank}\:{you}\:{very}\:{much}\: \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Mar/19
$${let}\:\left(\mathrm{100}{n}\right){transistor} \\ $$$${type}\:{A}=\mathrm{30}{n} \\ $$$${type}\:{B}=\mathrm{70}{n} \\ $$$$\left.{i}\right)\frac{\mathrm{30}{n}_{{c}_{\mathrm{4}} } }{\mathrm{100}{n}_{{c}_{\mathrm{4}} } }=\frac{\frac{\mathrm{30}{n}!}{\mathrm{4}!\left(\mathrm{30}{n}−\mathrm{4}\right)!}}{\frac{\mathrm{100}{n}!}{\left(\mathrm{100}{n}−\mathrm{4}\right)!\mathrm{4}!}}=\frac{\mathrm{30}{n}!×\left(\mathrm{100}{n}−\mathrm{4}\right)!}{\left(\mathrm{30}{n}−\mathrm{4}\right)!×\mathrm{100}{n}!} \\ $$$$=\frac{\mathrm{30}{n}×\left(\mathrm{30}{n}−\mathrm{1}\right)×\left(\mathrm{30}{n}−\mathrm{2}\right)\left(\mathrm{30}{n}−\mathrm{3}\right)}{\mathrm{100}{n}×\left(\mathrm{100}{n}−\mathrm{1}\right)×\left(\mathrm{100}{n}−\mathrm{2}\right)\left(\mathrm{100}{n}−\mathrm{3}\right)} \\ $$$$=\frac{\mathrm{3}}{\mathrm{10}}×\frac{\mathrm{30}−\frac{\mathrm{1}}{{n}}}{\mathrm{100}−\frac{\mathrm{1}}{{n}}}×\frac{\mathrm{30}−\frac{\mathrm{2}}{{n}}}{\mathrm{100}−\frac{\mathrm{2}}{{n}}}×\frac{\mathrm{30}−\frac{\mathrm{3}}{{n}}}{\mathrm{100}−\frac{\mathrm{3}}{{n}}} \\ $$$$\approx\frac{\mathrm{3}}{\mathrm{10}}×\frac{\mathrm{30}}{\mathrm{100}}×\frac{\mathrm{30}}{\mathrm{100}}×\frac{\mathrm{30}}{\mathrm{100}} \\ $$$$=\left(\frac{\mathrm{3}}{\mathrm{10}}\right)^{\mathrm{4}} {pls}\:{check} \\ $$$$\left.{ii}\right){for}\:{B}\:{type}\:=\left(\frac{\mathrm{7}}{\mathrm{10}}\right)^{\mathrm{4}} \: \\ $$$${pls}\:{check}… \\ $$
Commented by byaw last updated on 03/Mar/19
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{bro} \\ $$