A box contains 12 resistors, three of which are defective. If four are selected, find the probability of getting one defective resistor.
prob(defective) = 3/12 = 1/4
prob(not defective) = 3/4
prob(getting 1 defective when 4 selected)
= C(4,1)(1/4)(3/4)^3
= 4(1/4)(27/64)
= 27/64
To find the probability of getting one defective resistor, we need to find the total number of ways to select four resistors and the number of ways to select one defective resistor.
Total number of ways to select four resistors from a box of 12:
The number of ways to select four resistors from 12 is given by the combination formula. It can be calculated as C(12, 4) which is equal to 12! / (4! * (12 - 4)!) = 12! / (4! * 8!) = (12 * 11 * 10 * 9) / (4 * 3 * 2 * 1) = 495.
Number of ways to select one defective resistor from three:
Since three resistors are defective, the number of ways to select one defective resistor out of three is C(3, 1) = 3.
Therefore, the probability of getting one defective resistor out of the selected four resistors is:
Number of ways to select one defective resistor / Total number of ways to select four resistors = 3/495 = 1/165.
Hence, the probability of getting one defective resistor is 1/165.
To find the probability of getting one defective resistor out of four selected resistors, we need to calculate the following:
Number of ways to select one defective resistor out of three defective resistors: C(3, 1) = 3
Number of ways to select three working resistors out of the nine working resistors: C(9, 3) = 84
Number of total ways to select four resistors out of twelve: C(12, 4) = 495
The probability of selecting one defective resistor is given by:
P(one defective) = (Number of favorable outcomes) / (Number of total outcomes)
= (Number of ways to select one defective resistor) / (Number of total ways to select four resistors)
= 3 / 495
Thus, the probability of getting one defective resistor out of four selected resistors is 3/495 or simplified as 1/165.