A manufacturer ship toasters in cartons of 10. In each carton, they estimate a 20% chance that one of the toasters will need to be sent back for minor repairs. What is the probability, that in one carton there will be 5 toasters that need repair?
I have no idea on where to begin to answer the question above...
from table
0.026
What is the C stand for
Well, let's toast this question and see if we can come up with an answer!
If the manufacturer estimates a 20% chance that one toaster will need repairs, we can use this information to calculate the probability of having exactly 5 toasters in need of repairs in one carton.
To do that, we need to use a concept called the binomial distribution. The formula for the binomial distribution is P(X = k) = C(n, k) * p^k * (1-p)^(n-k), where:
- P(X = k) is the probability of having exactly k successes,
- C(n, k) is the number of ways to choose k successes out of n trials,
- p is the probability of success in one trial,
- (1-p) is the probability of failure in one trial,
- n is the number of trials.
In this case, n is the number of toasters in one carton (10), k is the number of toasters in need of repairs (5), and p is the probability that one toaster will need repairs (20%, which can be written as 0.2).
Now, let's plug these values into the binomial distribution formula:
P(X = 5) = C(10, 5) * 0.2^5 * (1-0.2)^(10-5)
Calculating C(10, 5) (which represents the number of ways to choose 5 out of 10) gives us 252:
P(X = 5) = 252 * 0.2^5 * 0.8^5
And, doing the math:
P(X = 5) ≈ 0.0264
So, there's approximately a 2.64% chance of having exactly 5 toasters in need of repair in one carton. Keep in mind that this is just an estimate based on the manufacturer's probability estimate. But hey, at least you can use this information to toast some statistics at your next toaster party!
To calculate the probability of having exactly 5 toasters that need repair in one carton, we need to use the binomial probability formula.
The formula for binomial probability is:
P(X=k) = (n C k) * (p^k) * ((1-p)^(n-k))
Where:
P(X=k) is the probability of having exactly k successes,
(n C k) is the binomial coefficient, also known as "n choose k",
p is the probability of success for a single trial, and
n is the number of trials.
For this problem, the number of trials (n) is the total number of toasters in a carton, which is 10. The probability of success (p) is the likelihood that a toaster needs repair, which is 20% or 0.2. We want to find the probability of getting exactly 5 toasters that need repair (k = 5).
So, applying the formula:
P(X=5) = (10 C 5) * (0.2^5) * ((1-0.2)^(10-5))
Now, let's calculate the individual components:
(10 C 5) = (10!)/((5!)(10-5)!) = (10*9*8*7*6)/(5*4*3*2*1) = 252
(0.2^5) = 0.2 * 0.2 * 0.2 * 0.2 * 0.2 = 0.00032
((1-0.2)^(10-5)) = 0.8^5 = 0.32768
Substituting the values back into the formula:
P(X=5) = 252 * 0.00032 * 0.32768
P(X=5) ≈ 0.0262144
Therefore, the probability that in one carton there will be 5 toasters that need repair is approximately 0.0262 or 2.62%.
Try the binomial probability table:
n = 10
x = 5
p = .20
q = 1 - p = 1 - .20 = .80
You can use the table, or calculate by hand using the following formula:
P(x) = (nCx)(p^x)[q^(n-x)]
With your data using the formula:
P(5) = (10C5)(.20^5)[.80^(10-5)]
I'll let you take it from here.