help please! in a sample of 26 hand held calculators, 20 are known to be non functional. if 6 of these calculators are selected at random, what is the probability that exactly four in the selection are non functional?

Question

campbell_st · Answer

looks like binomial probability    p = 20/26 and q = 6/26

P(non function = 4) = 6C4 (20/26)^2  x (6/26)^4

anonymous · Answer

well the options are a. .667, b. .316, c. .769, d. .300, and e. 0
i didn't get any of those when i did the equation