Showing Independent Cases in Binomial Distribution

Question

Can someone please give me a hint on how to solve part 2 in the following question? I am not sure how to even begin...

A professional proofreader has a $98 \%$ chance of detecting an error in a piece of written work (other than misspellings, double words, and similar errors that are machine detected). A work contains four errors.

1. Find the probability that the proofreader will miss at least one of them.
2. Show that two such proofreaders working independently have a $99.96\%$ chance of detecting an error in a piece of written work.

(Original image here.)

Answer 1

The probability that the first proofreader will miss the error is $0.02$. Same probability has the second proofreader. The complementary event of the event of interest is "both will miss the error". Due to independence, the probability of such event is multiplication of the two probabilities of missing the error, i.e., $0.02^2$, hence your answer is $$ 1-0.02^2=0.9996 . $$

Answer 2

Suppose both proof readers missed the error the probability of this event is$$ = \frac{2}{100}\cdot \frac{2}{100}=\frac{4}{10000}=0.04 \% $$

thus the probability of both working independently and detecting the error is $100-0.04=99.96\%$