the failure rate is .5%.
this means that the probability of a unit failing is .005
this means that the probability of a unit not failing is .995
the probability that a shipment has more than 2 defective batteries is equal to 1 minus the probability that it has less than or equal to 2 defective batteries.
the formula for determining the probability of exactly x defective batteries is:
p(x) = C(n,x) * p(f)^x * p(s)^(n-x)
x = number of defective batteries
n = total number of batteries
p(f) = probability of failure = .005
p(s) = probability of success = 1 - .005 = .995
C(n,x) = the combination of x things out of a universe of n things.
an example will clarify how this works.
suppose the shipment only contained 3 batteries.
x = number of failures.
we'll replace x with the number of failures to be analyzed.
p(0) = C(n,0) * .005^0 * .995^3 = 1 * 1 * .995^3 = .985074875
P(1) = C(n,1) * .005^1 * .995^2 = 3 * .005^1 * .995^2 = .014850375
p(2) = C(n,2) * .005^2 * .995^1 = 3 * .005^2 * .995^1 = .000074625
p(3) = C(n,3) * .005^3 * .995^0 = 1 * .005^3 * .995^0 = .000000125
add these 4 probabilities together and you get 1 which is the total of all probabilities, as it should be.
now, if i was asking for the probability that more than 1 batteries failed, that would be equal to the probability of 2 failing and 3 failing which would be equal to .000074625 + .000000125 = .00007475.
that would also be equal to 1 minus the probability of 0 failing and 1 failing which would be equal to 1 - .985074875 - .014850375 which would be equal to .00007475.
we get the same number even though we approached the problem from a different perspective.
when n = 1000, the same concepts apply except the direct approach is way too calculation intensive.
to directly analyze it, we would have to take p(3) + p(4) + p(1000).
no way we want to do that if we can avoid it.
so we take the alternate approach.
probability of more than 2 defective batteries equals 1 minus p(0) - p(1) - p(2).
a word about C(n,x).
C(n,x) = n! / (x! * (n-x)!)
to put this into numbers, suppose n = 5 and x = 2.
then C(n,x) = C(5,2) = 5! / (2! * 3!)
this would be equal to (5 * 4 * 3 * 2 * 1) / ((2 * 1) * (3 * 2 * 1))
you can see that the (3 * 2 * 1) in the numerator and the denominator cancel out and you are left with (5 * 4) / (2 * 1) which then becomes (5 * 2) which equals 10.
the same concept is applied to your problem, but the numbers are much larger.
in your problem
n = 1000
x will be equal to 0, 1, or 2
p(f) = .005
p(s) = .995
your answer will be 1 - p(0) - p(1) - p(2) which will give you the probability that more than 2 batteries are defective.
the formulas that you will use are:
p(0) = C(1000,0) * .005^0 * .995^1000
p(1) = C(1000,1) * .005^1 * .995^999
p(2) = C(1000,2) * .005^2 * .995^998
C(1000,0) = 1000! / (0! * 1000!)
since 0! = 1, this becomes:
C(1000,0) = 1000! / 1000! which equals 1
C(1000,1) = 1000! / (1! * 999!)
this is equivalent to (1000 * 999!) / (1 * 999!).
the 999! in the numerator and denominator cancel out and you are left with:
C(1000,1) = 1000 / 1 which equals 1000
C(1000,2) = 1000! / (2! * 998!)
this is equivalent to (1000 * 999 * 998!) / (2 * 1 * 998!)
the 998! in the numerator and denominator cancel out and you are left with:
C(1000,2) = (1000 * 999) / 2 which equals 499500.
the formulas that you will use are:
p(0) = C(1000,0) * .005^0 * .995^1000
p(1) = C(1000,1) * .005^1 * .995^999
p(2) = C(1000,2) * .005^2 * .995^998
after replacing the calculations for C(n,x) we just accomplished above, these formulas become:
p(0) = 1 * .005^0 * .995^1000
p(1) = 1000 * .005^1 * .995^999
p(2) = 499500 * .005^2 * .995^998
we multiply these out to get:
p(0) = .006653969
p(1) = .033437028
p(2) = .083928621
p(0) + p(1) = p(2) = .124019617
1 - p(0) - p(1) - p(2) = 1 - .124019617 = .875980383
the probability that more than 2 batteries are defective is .875980383.
this appears to be a reasonable answer considering that the failure rate is .005.
if this failure rate was derived from experience, then an average of approximately 5 out of every 1000 batteries that were manufactured must have been defective.
since 2 is less than 5, the probability that more than 2 will be defective should be pretty high, as it is.
Step-by-step explanation:
