A study shows that 6% of household appliances fail during any given year. If a person has 6 household appliances, what is the probability that exactly 2 of them will fail next year?
(9/2)*.06^2*.94^7 this is how I worked it but it was wrong can you tell me where I went wrong so that I can have a good understanding of how to work problem correctly.
To find the probability of exactly 2 out of 6 household appliances failing next year, you can use the binomial probability formula. The formula is as follows:
P(x) = (nCx) * p^x * (1-p)^(n-x)
Where:
P(x) is the probability of getting exactly x successes
n is the total number of trials (in this case, the number of household appliances = 6)
x is the number of successes (in this case, exactly 2 of them failing)
p is the probability of success for each individual trial (in this case, the probability of an appliance failing = 0.06)
Let's calculate it step by step:
Step 1: Calculate the number of ways to choose exactly 2 appliances out of 6.
nCx = n! / (x! * (n-x)!)
= 6! / (2! * (6-2)!)
= 6! / (2! * 4!)
= (6*5) / (2*1) = 15
Step 2: Raise the probability of failure (0.06) to the power of the number of appliances failing (2).
p^x = 0.06^2 = 0.0036
Step 3: Raise the complement of the failure probability (1 - 0.06) to the power of the number of appliances not failing (6-2 = 4).
(1-p)^(n-x) = (1-0.06)^4 = 0.94^4 = 0.78408
Step 4: Multiply all the values together to get the final probability.
P(2 appliances failing) = (nCx) * p^x * (1-p)^(n-x)
= 15 * 0.0036 * 0.78408
= 0.0419 (rounded to four decimal places)
So, the probability of exactly 2 out of 6 household appliances failing next year is approximately 0.0419 or 4.19%.