Question #179600

A secretary makes two error per page on the average. Wht is the probability that on the next page she makes. a) At least 3 errors? b) 6 to 8 errors?


1
Expert's answer
2021-04-15T07:25:25-0400

We use Poisson's formula p(m)=λmm!eλp(m) = \frac{{{\lambda ^m}}}{{m!}}{e^{ - \lambda }} . In our case λ=2\lambda = 2 .

a) Let's find the probabilities that 0, 1 and 2 errors will be made:

p(0)=200!e2=e2;p(1)=211!e2=2e2;p(2)=222!e2=2e2p(0) = \frac{{{2^0}}}{{0!}}{e^{ - 2}} = {e^{ - 2}};\,\,p(1) = \frac{{{2^1}}}{{1!}}{e^{ - 2}} = 2{e^{ - 2}};\,p(2) = \frac{{{2^2}}}{{2!}}{e^{ - 2}} = 2{e^{ - 2}}

Then the probability that the secretary makes less than 3 errors is

p(x<3)=p(0)+p(1)+p(2)=5e2p(x < 3) = p(0) + p(1) + p(2) = 5{e^{ - 2}}

Whence the wanted probability, as the probability of the opposite event, is

p(x3)=1p(x<3)=15e20.323p(x \ge 3) = 1 - p(x < 3) = 1 - 5{e^{ - 2}} \approx 0.323

Answer: p(x3)0.323p(x \ge 3) \approx 0.323

b) Let's find the probabilities that 6, 7 and 8 errors will be made:

p(6)=266!e2=445e2;p(7)=277!e2=8315e2;p(8)=288!e2=2315e2p(6) = \frac{{{2^6}}}{{6!}}{e^{ - 2}} = \frac{4}{{45}}{e^{ - 2}};\,\,p(7) = \frac{{{2^7}}}{{7!}}{e^{ - 2}} = \frac{8}{{315}}{e^{ - 2}};\,p(8) = \frac{{{2^8}}}{{8!}}{e^{ - 2}} = \frac{2}{{315}}{e^{ - 2}}

Then wanted probability is

p(6x8)=p(6)+p(7)+p(8)=38315e20.0163p(6 \le x \le 8) = p(6) + p(7) + p(8) = \frac{{38}}{{315}}{e^{ - 2}} \approx 0.0163

Answer: p(6x8)0.0163p(6 \le x \le 8) \approx 0.0163


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