Answer to Question #179600 in Statistics and Probability for kainat

Question

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?

Answer 1

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

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

$p(0) = \frac{{{2^0}}}{{0!}}{e^{ - 2}} = {e^{ - 2}};\,\,p(1) = \frac{{{2^1}}}{{1!}}{e^{ - 2}} = 2{e^{ - 2}};\,p(2) = \frac{{{2^2}}}{{2!}}{e^{ - 2}} = 2{e^{ - 2}}$

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

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

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

$p(x \ge 3) = 1 - p(x < 3) = 1 - 5{e^{ - 2}} \approx 0.323$

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

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

$p(6) = \frac{{{2^6}}}{{6!}}{e^{ - 2}} = \frac{4}{{45}}{e^{ - 2}};\,\,p(7) = \frac{{{2^7}}}{{7!}}{e^{ - 2}} = \frac{8}{{315}}{e^{ - 2}};\,p(8) = \frac{{{2^8}}}{{8!}}{e^{ - 2}} = \frac{2}{{315}}{e^{ - 2}}$

Then wanted probability is

$p(6 \le x \le 8) = p(6) + p(7) + p(8) = \frac{{38}}{{315}}{e^{ - 2}} \approx 0.0163$

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