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
a record company claims that of 55 000 cd's produced 15% will be defective, 100 were defective. is the company right?
To find the probability of getting no defective resistors, one defective resistor, and three defective resistors when four resistors are selected from the package, we need to use the concept of combinations.
First, let's calculate the total number of ways to select four resistors from the package. This can be found using the combination formula:
nCr = n! / (r!(n-r)!)
where n is the total number of resistors in the package (12 in this case) and r is the number of resistors being selected (4 in this case).
So, the total number of ways to select four resistors from a package of 12 resistors is:
12C4 = 12! / (4!(12-4)!) = 12! / (4!8!) = (12 * 11 * 10 * 9) / (4 * 3 * 2 * 1) = 495
Now, let's calculate the probability of getting no defective resistors:
To get no defective resistors, we need to select four resistors from the remaining nine non-defective resistors in the package. So, the number of favorable outcomes is:
9C4 = 9! / (4!(9-4)!) = 9! / (4!5!) = (9 * 8 * 7 * 6) / (4 * 3 * 2 * 1) = 126
The probability of getting no defective resistors is then:
P(No defective resistors) = favorable outcomes / total outcomes = 126 / 495 = 0.2545 (rounded to four decimal places)
Next, let's calculate the probability of getting one defective resistor:
To get one defective resistor, we can choose one defective resistor from the three defective ones available and then select three non-defective resistors from the remaining nine non-defective ones. So, the number of favorable outcomes is:
3C1 * 9C3 = (3! / (1!(3-1)!)) * (9! / (3!(9-3)!) = (3 * 2 / 1 * 2) * (9 * 8 * 7 / (3 * 2 * 1)) = 3 * 84 = 252
The probability of getting one defective resistor is then:
P(One defective resistors) = favorable outcomes / total outcomes = 252 / 495 = 0.5091 (rounded to four decimal places)
Finally, let's calculate the probability of getting three defective resistors:
To get three defective resistors, we can choose three defective resistors from the three defective ones available and then select one non-defective resistor from the remaining nine non-defective ones. So, the number of favorable outcomes is:
3C3 * 9C1 = (3! / (3!(3-3)!)) * (9! / (1!(9-1)!)) = (1 / 1) * (9 / 1) = 1 * 9 = 9
The probability of getting three defective resistors is then:
P(Three defective resistors) = favorable outcomes / total outcomes = 9 / 495 = 0.0182 (rounded to four decimal places)
Therefore, the requested probabilities are:
- Probability of getting no defective resistors: 0.2545
- Probability of getting one defective resistor: 0.5091
- Probability of getting three defective resistors: 0.0182
Assuming no replacement,
The probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
None defective = 9/12 * 8/11 * 7/10 * 6/9 = ?
One defective = 3/12 * 9/11 * 8/10 * 7/9 = ?
Three defective = 3/12 * 2/11 * 1/10 * 9/9 = ?