you have to find the number of ways to get four people without there being any couples. The number of permutations for this is going to be

10 * 8 * 6 * 4 * 2 = 3848

where you have 10 people that could be the first choice then only 8 left for the second as one has been picked and their spouse is also not an option. then you have only six left, the two picked and their spouse are removed from the options list. and so on.

the number of permutations you can have for the ten people is 10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 * 0! where 0! = 1 by definition.

10! = 3628800

the probability is: 3840 / 3628800 = 1/945 = 0.00106

—–

the lots have 10% defective chips. Let X be the number of defective chips in a four chip sample. X has the binomial distribution with 4 trials and success probability of 0.10.

In general, if X has the binomial distribution with n trials and a success probability of p then

P[X = x] = n!/(n!(n-1)!) * p^x * (1-p)^(n-x)

for values of x = 0, 1, 2, …, n

P[X ≥ 1] = 1 – P[X = 0]

P[X = 0] = 0.6561

P[X ≥ 1] = 1 – P[X = 0] = 0.3439