A certain type of battery has a .5% failure rate. Find the probability that a shipment of 1000 batteries has more than two defective batteries.

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:

The answer is B.

Step-by-step explanation:

You can see that the triangles are identical, so find the side that matches with BA and there you have it. The corresponding side is ED. I hope this helps!