A factory makes 10% defective items and items are independently defective. If a sample of 10 items is to be selected, find the probability that 9 or more are NOT defective in two ways. (Round to 3 decimal places)
a. Let "Success" refer to a non-defective item and use X = number of good items and the binomial distribution.
Probability =
b. Let "Success" refer to a defective item and use X = number of bad items and the binomial distribution. Probability =
Wouldn't the probability be 1 for both of these questions?
a)
Let $X$ be the discrete random variable that expresses the number of the non-defective items in our sample. Assuming that the probability of getting a non-defective item remains the same, the probability of success is $p=1-0.1 =0.9$ Now, considering the formula of the binomial distribution (with $p=0.9$ and $n=10$), we have: $$\Pr(X \ge 9)\begin{array}[t]{l}=\Pr(X=9) +\Pr(X=10)=\sum\limits_{k=9}^{n}\binom{n}{k}\cdot p^k \cdot (1-p)^{n-k}\\ =\sum\limits_{k=9}^{10}\binom{10}{k}\cdot 0.9^k\cdot 0.1^{10-k}\\ \approx 0.736 \end{array}$$
For b) you may work in a similar way.