A manufacturer of light bulbs knows that 3% of the production of their 60W bulbs will be defective. What is the probability that exactly 5 bulbs in a carton of 144 will ne defective?
(0.03)^5*(0.97)^139*144!/(139!*5!)
= 2.43*10^-8*1.45*10^2*5.77*10^1/120
= 0.169
To find the probability that exactly 5 bulbs in a carton of 144 will be defective, we can use the binomial probability formula.
The formula for the probability of getting exactly k successes (defective bulbs) in n trials (total bulbs) with a probability of success (probability of a bulb being defective) p is:
P(X = k) = (nCk) * (p^k) * ((1-p)^(n-k))
In this case, n = 144 (total bulbs), k = 5 (defective bulbs), and p = 0.03 (probability of a bulb being defective).
Using the combination formula nCk = n! / (k! * (n-k)!), we can calculate the binomial probability:
P(X = 5) = (144C5) * (0.03^5) * ((1-0.03)^(144-5))
Calculating this expression gives us:
P(X = 5) = (144! / (5! * (144-5)!)) * (0.03^5) * (0.97^139)
P(X = 5) ≈ 0.284
Therefore, the probability that exactly 5 bulbs in a carton of 144 will be defective is approximately 0.284, or 28.4%.
To find the probability that exactly 5 bulbs in a carton of 144 will be defective, we can use the binomial probability formula. The formula is:
P(X = k) = nCk * p^k * (1-p)^(n-k)
where:
- P(X = k) is the probability of exactly k successes
- nCk is the number of combinations of n items taken k at a time
- p is the probability of success
- (1-p) is the probability of failure
- n is the total number of trials
- k is the number of successful trials
In this case:
- p = 0.03 (the probability of a bulb being defective)
- n = 144 (total number of bulbs in the carton)
- k = 5 (number of defective bulbs we want to find the probability for)
Now let's substitute these values into the formula:
P(X = 5) = 144C5 * 0.03^5 * (1-0.03)^(144-5)
To calculate the values in the formula:
- 144C5 can be calculated as 144! / (5! * (144-5)!). Here, "!" represents the factorial function.
- 0.03^5 can be evaluated as 0.03 * 0.03 * 0.03 * 0.03 * 0.03.
- (1-0.03)^(144-5) can be evaluated as 0.97^139.
By substituting these values and calculating, we can find the probability that exactly 5 bulbs in a carton of 144 will be defective.