Assuming that 10% is the probability of a switch being defective (since left-handedness has nothing to do with the problem):
For a binomial distribution, probability(r out of n) = (nCr) (p)^r (q)^(1-r)
p = 10% = 0.1, q = 1 - p = 0.9
r = 2 switches, n = total of 12 switches
probability = (12C2) (0.1)^2 (0.9)^(12-2)
probability = 66(0.1)^2 (0.9)^10
probability = 0.23
