a study shows that 16% of household appliances break during any given year. If a person has 4 household appliances, what is the probability that exactly 1 of them will break next year? I just need a formula and I can get it from there.
would that not be
.16 * (1-.16)^3
?
Steve no that does not produce the correct answer
hmmm. The only other candidate I see would be 4 times that. You know
P(only 1st) + P(only 2nd) + ...
To calculate the probability that exactly one household appliance will break next year, you can use the binomial probability formula:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
Where:
P(X = k) is the probability of getting exactly k successes (in this case, exactly 1 appliance breaking)
n is the total number of trials (in this case, the number of appliances)
k is the number of desired successes (in this case, 1 appliance breaking)
p is the probability of success on a single trial (in this case, the probability of an appliance breaking - 0.16)
C(n, k) is the binomial coefficient, calculated as C(n, k) = n! / (k! * (n-k)!)
So, plugging in the values:
P(X = 1) = C(4, 1) * 0.16^1 * (1-0.16)^(4-1)
Calculating C(4, 1) = 4! / (1! * (4-1)!) = 4! / (1! * 3!) = (4 * 3 * 2) / (1 *2 * 1) = 4
P(X = 1) = 4 * 0.16 * (1-0.16)^3
Calculating (1-0.16)^3 = 0.84^3 = 0.592704
P(X = 1) = 4 * 0.16 * 0.592704
P(X = 1) = 0.378112
Therefore, the probability that exactly one household appliance will break next year is approximately 0.378112 or 37.81%.