Combinations and Probability with Keyboards

Combinations and Probability with Keyboards
$\begingroup$

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!

$\endgroup$
1

2 Answers

$\begingroup$

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?

$$P = \frac{{19\choose 4}+{19\choose 5}}{25 \choose 5}$$

$\endgroup$
1
$\begingroup$

$\newcommand{\ch}[2]{{^{#1}\mathsf C_{#2}}}$

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. $$\dfrac{\ch{19}4 \cdot \ch{6}{1} + \ch{19}5}{\ch{25}5}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Alexander Ross
Author

Alexander Ross

Alexander Ross has covered the video game industry for a decade, writing deep dives on game design, esports tournaments, VR developments, and gaming culture.