From a box of 12 flares, four are selected at random. If the box contains four flares that do not work:
a) what is the probability that all four will not work?
b) what is the probability that, at most, two will not work?
c) of the four flares selected, how many would you expect not to work?
+1223 There are 4 nonworking flares and 8 working flares.
a) \(\frac{4 \choose 4}{12 \choose 4} = \boxed{\frac{1}{495}}\)
b) \(\frac{4 \choose 4}{12 \choose 4} + \frac{\binom{4}{3} \cdot \binom{8}{1}}{12 \choose 4} + \frac{\binom{4}{2} \cdot \binom{8}{2}}{12 \choose 4}= \boxed{\frac{67}{165}}\)
c) \(\frac{\binom{4}{4}}{\binom{12}{4}}(0) + \frac{\binom{4}{3} \cdot \binom{8}{1}}{\binom{12}{4}}(1) + \frac{\binom{4}{2} \cdot \binom{8}{2}}{\binom{12}{4}}(2) + \frac{\binom{4}{1} \cdot \binom{8}{3}}{\binom{12}{4}}(3) + \frac{\binom{8}{4}}{\binom{12}{4}}(4) = \frac{1320}{495} = \boxed{\frac{8}{3}} \)
