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Question Number 55737 by byaw last updated on 03/Mar/19

Commented by mr W last updated on 03/Mar/19

$$\left(i\right)$$$$p={(0.3)}^4=0.0081$$$$\left(ii\right)$$$$p={(0.7)}^4=0.2401$$

Commented by byaw last updated on 03/Mar/19

$$Thankyouverymuch$$

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Mar/19

$$let\left(100n\right)transistor$$$$typeA=30n$$$$typeB=70n$$$$i)\frac{30n_{c_4}}{100n_{c_4}}=\frac{\frac{30n!}{4!(30n-4)!}}{\frac{100n!}{(100n-4)!4!}}=\frac{30n!\times (100n-4)!}{(30n-4)!\times 100n!}$$$$=\frac{30n\times (30n-1)\times (30n-2)(30n-3)}{100n\times (100n-1)\times (100n-2)(100n-3)}$$$$=\frac{3}{10}\times \frac{30-\frac{1}{n}}{100-\frac{1}{n}}\times \frac{30-\frac{2}{n}}{100-\frac{2}{n}}\times \frac{30-\frac{3}{n}}{100-\frac{3}{n}}$$$$\approx \frac{3}{10}\times \frac{30}{100}\times \frac{30}{100}\times \frac{30}{100}$$$$={\left(\frac{3}{10}\right)}^{4}plscheck$$$$ii)forBtype={\left(\frac{7}{10}\right)}^4$$$$plscheck...$$

Commented by byaw last updated on 03/Mar/19

$$\mathrm{Thank}\mathrm{you}\mathrm{bro}$$

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