When purchasing bulk orders of​ batteries

Question

When purchasing bulk orders of​ batteries, a toy manufacturer uses this acceptance sampling​ plan: Randomly select and test 42 batteries and determine whether each is within specifications. The entire shipment is accepted if at most 3 batteries do not meet specifications. A shipment contains 3000

​batteries, and 2​% of them do not meet specifications. What is the probability that this whole shipment will be​ accepted?

Will almost all such shipments be​ accepted, or will many be​ rejected?

Answer 1

Hi Amanda,

This is a binomial probability problem. We have two potential outcomes--defective battery or not. Formula for that is:

P=C(n,x)p^xq^n-x

Breaking this down:

C implies combination--order in which you select the batteries does not matter

n=sample size

x=desired number of "successes" i.e. defective batteries

p=probability of success

q=1-p=probability of failure

Now, we have four separate calculations to do because the question asked "At most, 3 defective batteries." That means we have to compute probabilities of 3,2,1 and 0 defectives and then add them up.

I'll start from the top:

P=C(n,x)p^xq^n-x

n=42

x=3

p=0.02

q=1-p=0.98

So:

P=C(42,3)0.02³*0.98³⁹

Now, the combination C(42,3) has its own formula:

C(n,x)=n!/x!(n-x)!

where:

n=sample size

x=desired number of successes

!=factorial=n*(n-1)*(n-2)...i.e. 3*2*1

So, substituting and calculating:

C(42,3)=42!/(3!*39!)=11480

Now, back to the original problem:

P=C(42,3)0.02³*0.98³⁹

C(42,3)=11480

I will use P₃ so you know which x I am calculating. I'll use P_2, P₁, P₀ for the rest. Thus:

P₃=11480*0.02³*0.98³⁹

P₃=0.042

Keep this probability in mind, as we will need it at the end.

Now, we need P₂.

P₂=C(42,2)*0.02²*0.98⁴⁰

We need to compute C(42,2) with the factorial formula:

C(42,2)=42!/(2!*40!)=861

So:

P₂=861*0.02²*0.98⁴⁰

P₂=0.153

Let's get P₁

P₁=C(42,1)*0.02¹*0.98⁴¹

Factorial formula time:

C(42,1)=42!/(1!*41!)=42

Substituting:

P₁=42*0.02¹*0.98⁴¹

P₁=0.367

Now, let's get P₀:

P₀=C(42,0)*0.02⁰*0.98⁴²

Now, we are going to do the factorial formula again, but note that:

0!=1, by definition

Also, any number divided by itself is 1, so:

C(42,0)=42!/(42!*0!)=1

Substituting:

P₀=1*0.02⁰*0.98⁴²

Additionally

a⁰=1, where a is any real number, so:

P₀=1*1*0.98⁴²

P₀=0.98⁴²

P₀=0.428

Now, we can add up the probabilities to get the probability that the whole shipment will be accepted. This is P(X<=3)

P(X<=3)=0.990

Therefore, almost all shipments (99%) will be accepted; only 1% will be rejected. I hope this helps.