Let x be the case that calculator Does function. x' means it will not function.
Then P(x') = 1 - P(x)
Its easier to find P(x) since it requires that both batteries are good.
There are 3 dead batteries and 5 good ones.
P(x) can be found by dividing at the combinations which gives 2 good batteries by total combinations.
\(P(x) = \frac{5C2}{8C2} = \frac{10}{28} = \frac{5}{14} \\ \\ P(x') = 1 - \frac{5}{14} = \frac{9}{14}\)
Therefore there is a 9/14 chance the calculator will not function.
