A package contains 12 resistors, 3 of which are defective. If four are selected, find the
probability of getting the following.
No defective resistors
One defective resistors
3 defective resistors
Nbn
Bbbb
To find the probability of getting a specific number of defective resistors when selecting four resistors from a package of 12, we can use the concept of combinations.
First, let's determine the total number of possible combinations of selecting four resistors from a total of 12. We can use the combination formula:
nCr = n! / (r! * (n - r)!)
Here, n represents the total number of items and r represents the number of items we are selecting.
For our case, n = 12 (total number of resistors) and r = 4 (number of resistors to be selected).
nCr = 12! / (4! * (12 - 4)!)
= 12! / (4! * 8!)
Calculating the above expression, we get:
nCr = (12 * 11 * 10 * 9) / (4 * 3 * 2 * 1)
= 495
So, there are 495 possible combinations when selecting four resistors from a total of 12.
Now, let's calculate the probabilities for each case:
1. No defective resistors:
To calculate the probability of selecting four resistors with no defective ones, we need to consider the good resistors only. In this case, there are 12 - 3 = 9 good resistors.
The number of combinations to select four good resistors out of nine is given by:
nCr = 9! / (4! * (9 - 4)!)
= 9! / (4! * 5!)
= (9 * 8 * 7 * 6) / (4 * 3 * 2 * 1)
= 126
So, there are 126 possible combinations with no defective resistors.
The probability is then calculated as:
P(No defective resistors) = Number of favorable outcomes / Total number of outcomes
= 126 / 495
= 14 / 55
Therefore, the probability of selecting four resistors with no defective ones is 14/55.
2. One defective resistor:
To calculate the probability of selecting one defective resistor, we need to consider the number of ways we can choose one defective resistor and three good resistors.
The number of combinations to select one defective resistor out of three is given by:
nCr = 3! / (1! * (3 - 1)!)
= 3
The number of combinations to select three good resistors out of nine (12 - 3) is given by:
nCr = 9! / (3! * (9 - 3)!)
= 9! / (3! * 6!)
= (9 * 8 * 7) / (3 * 2 * 1)
= 84
So, there are 3 * 84 = 252 possible combinations with one defective resistor.
The probability is then calculated as:
P(One defective resistor) = Number of favorable outcomes / Total number of outcomes
= 252 / 495
Therefore, the probability of selecting four resistors with one defective resistor is 252/495.
3. Three defective resistors:
To calculate the probability of selecting three defective resistors, we can directly use the number of combinations to select three defective resistors out of three, which is equal to 1.
The probability is then calculated as:
P(Three defective resistors) = Number of favorable outcomes / Total number of outcomes
= 1 / 495
Therefore, the probability of selecting four resistors with three defective ones is 1/495.