For every 1000 batteries produced by a manufacturer 15 are defective. What is the
probability that 100 randomly chosen batteries will have exactly 2 defective batteries?
d = defective , n = not defective
this is a binomial distribution
... a battery is or isn't
100C2 * .985^98 * .015^2
To find the probability that 100 randomly chosen batteries will have exactly 2 defective batteries, we can use the binomial probability formula.
The binomial probability formula is given by:
P(X=k) = (nCk) * p^k * (1-p)^(n-k)
Where:
- P(X=k) is the probability of getting k successes (in this case, 2 defective batteries).
- n is the total number of trials (in this case, 100 batteries).
- k is the number of successes (in this case, 2 defective batteries).
- p is the probability of success (in this case, the probability of a battery being defective).
- (nCk) is the combination formula that calculates the number of ways to choose k items from a set of n items.
Given that 15 out of every 1000 batteries are defective, the probability of a battery being defective is 15/1000, or 0.015.
Now, let's substitute these values into the formula:
P(X=2) = (100C2) * (0.015)^2 * (1-0.015)^(100-2)
To calculate (100C2), which represents the number of ways to choose 2 out of 100 batteries, we use the combination formula:
(100C2) = 100! / (2! * (100-2)!)
The exclamation point denotes the factorial function, which means multiplying a number by all the positive integers less than it down to 1.
So, to find the probability of exactly 2 defective batteries out of 100 randomly chosen batteries produced by the manufacturer, you can follow these steps:
1. Calculate (100C2) using the combination formula.
2. Raise the probability of a battery being defective (0.015) to the power of 2.
3. Raise the probability of a battery NOT being defective (1 - 0.015) to the power of (100-2).
4. Multiply all three values together.
The result will give you the probability of having exactly 2 defective batteries out of 100 randomly chosen batteries.