A box contains 10 transistors, 4 of which are defective. If 4 are selected at random, find the probability of the statements below.
a.all are defective
B. None are defective
a. To find the probability that all 4 transistors selected are defective, we need to consider the number of ways we can select 4 defective transistors from the 4 defective transistors in the box, divided by the total number of ways we can select any 4 transistors from the 10 transistors in the box.
The number of ways to select 4 defective transistors from the 4 defective transistors in the box is 1 (since there is only one combination that satisfies this condition).
The total number of ways to select any 4 transistors from the 10 transistors in the box is given by the combination formula: C(10, 4) = 10! / (4! * (10-4)!) = 210.
Therefore, the probability that all 4 selected transistors are defective is 1/210.
b. To find the probability that none of the 4 selected transistors are defective, we need to consider the number of ways we can select 4 non-defective transistors from the 6 non-defective transistors in the box, divided by the total number of ways we can select any 4 transistors from the 10 transistors in the box.
The number of ways to select 4 non-defective transistors from the 6 non-defective transistors in the box is given by the combination formula: C(6, 4) = 6! / (4! * (6-4)!) = 15.
The total number of ways to select any 4 transistors from the 10 transistors in the box is again given by the combination formula: C(10, 4) = 210.
Therefore, the probability that none of the 4 selected transistors are defective is 15/210, which simplifies to 1/14.