The expected number of typographical errors on a page of a certain magazine is .2. What is the probability that the next page you read contains (a) 0 and (b) 2 or more typographical errors? Explain your reasoning!

The expected

hansel

Answered by Aaron 2 years ago

These circumstances fit a Poisson distribution,

$$P(X=k)=\frac{\lambda^k e^{-\lambda}}{k!}$$

with

$$\lambda = 0.2$$

errors per page.

**part (a)**

the probability that the next page contains 0 errors is given by,

$$P(X=0)=\frac{(0.2)^0e^{-0.2}}{0!}=e^{-0.2}$$

**part (b)**

the probability of two or more errors is given by,

$$P(X \ge 2)=1 - P(X < 2) =1-P(X=1)-P(X=0)$$

$$=1-\frac{(0.2)^1e^{-0.2}}{1!} - e^{-0.2}=1-0.2e^{-0.2}-e^{-0.2}\approx0.0175$$

Surround your text in `*italics*`

or `**bold**`

, to write a math equation use, for example, `$x^2+2x+1=0$`

or `$$\beta^2-1=0$$`

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Asked: 2 years ago