I'm not getting this can someone explain plz and i'll be very thankful to .

Suppose that a day's production of 850 manufactured parts contains 50 parts that do not con- form to customer requirements. Two parts are selected at random, without replacement, from the batch. Let the random variable X equal the number of nonconforming parts in the sample. What is the cumulative distribution function of X? The question can be answered by first finding the probability mass function of X. 709 P(X = 0) = = 0.886 850 819 P(X = 1) = 2 : 850 849 P(X = 2) =0.003 850 849 Therefore. F(0) = P(X = 0) = 0,886 F(I) = P(X = 1) = 0.886 + 0.111 = 0.997 F(2) = P(X $ 2) = 1 The cumulative distribution function for this example is graphed in Fig. 3-4. Note that F(x) is defined for all & from -> and not only for 0, I, and 2