# Statistics - Combinations and Probability

I have the question

Computer keyboard failures can be attributed to electrical defects or mechanical defects. A repair facility currently has 25 failed keyboards, 6 of which have electrical defects and 19 of which have mechanical defects

A) In how many ways can a sample of 5 keyboards be selected so that exactly two have an electrical defect?

B) If a sample of 5 keyboards is random selected, what is the probability that at least 4 of these will have a mechanical defect?

For A - I know i have to use combinations, but i'm not sure what the numbers are.

Could anyone begin to point me in the right direction? Thanks!

• @MichaelHardy No, it isn't. "None" is not the complement of "at least 4". – Graham Kemp Oct 4 '16 at 23:08

$$P = \frac{{19\choose 4}+{19\choose 5}}{25 \choose 5}$$
• You need to select the electrical defected keyboards too.$$P=\dfrac{\binom {19} 4\binom 6 1+\binom{19}5\color{silver}{\binom 6 0}}{\binom {25}5}$$ – Graham Kemp Oct 4 '16 at 23:05
$\newcommand{\ch}[2]{{^{#1}\mathsf C_{#2}}}$
A. From the 6 keyboards that have electrical defects, choose 2. From the 19 keyboards that have mechanical defects, choose 3. $$\ch6 2 \cdot \ch {19}3$$
B. From the 19 keyboards that have mechanical defects, choose 4 then from 6 keyboards that have electrical defects, choose 1. Add it to: from the 19 keyboards that have mechanical defects, choose 5. All over from 25 keyboards that have defects, choose 5. $$\dfrac{\ch{19}4 \cdot \ch{6}{1} + \ch{19}5}{\ch{25}5}$$