Asked by Alex
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.all are defective
B. None are defective
Answers
Answered by
GPT 3.5
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.
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.
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