Given, on average a box contains 2 defective items.

a. Probability that a randomly selected box has 2 defective items 

P(r = 2) = (e^-2 *2^r )/r!  = (e^-2 *2² )/2! = (0.13534*4)/2 = 0.13534*2 = 0.27068

b. Probability that a randomly selected box has at most 2 defective items

      P(r ≤ 2) = ∑_r (e^-2 *2^r )/r!   {0 ≤ r ≤ 2}

      P(r ≤ 2) = e^-2 *{1 + 2 + 4/2}

=> P(r ≤ 2) = e^-2 *{1 + 2 + 2}

=> P(r ≤ 2) = e^-2 *5

=> P(r ≤ 2) = 0.13534 *5

=> P(r ≤ 2) = 0.6767