Question

A certain typing agency employs 2 typists. The average number of errors per article is 3 when typed by the first typist and 4.2 when typed by the second. If your article is equally likely to be typed by either typist, approximate the probability that it will have no errors.

Answer 1

Answer:

0.03

Step-by-step explanation:

For the purpose of the question we will employ the formulas used in Poisson Distribution

$P(x; u) = \frac{(e^{-u} ) (u^{x} )}{x!}$

Where

$P$ = Probability

$u$ = Average number of errors

$x$ = Required number of errors

From the given information we can extract the following information

Artist 1

$u_{1}$ = 3

$x_{1}$ = 0

Artist 2

$u_{2}$ = 4.2

$x_{2}$ = 0

Now since this is an OR condition, we find their individual probabilities and add them

$P_{1} (0; 3)*P(Artist 1) = \frac{(e^{-3} ) (3^{0} )}{0!} * \frac{1}{2}  = \frac{1}{2}e^-^3 \\ P_{2} (0; 4.2)*P(Artist 2) = \frac{(e^{-4.2} ) (4.2^{0} )}{0!}*\frac{1}{2} = \frac{1}{2}  e^-^4^.^2$

Adding these together

$\frac{1}{2} (e^-^3 + e^-^4^.^2) = 0.03$