# Math Help - Probability

1. ## Probability

Could I please ask someone for either help how to do this or tell me what exactly this problem is called so I can look up the formula. I promise after I am done w/this Stats course to NEVER enter your domain again, that being the domain of the Stats/Calc world. I have already done this type of problem quite a few times but I have so many notes, and they all look the same, I am getting totally confused.

A manufacturer of electronic appliances purchases small hybrid circuit boards from an outside vendor. The circuits are shipped in large lots, with several thousand circuit boards per lot. When each lot is received, a random sample of 30 of the circuit boards is selected and tested to see if the circuitry performs correctly. If more than 5 of the circuit boards in the sample fail the testing, then the manufacturer rejects the entire lot and ships it back to the vendor. Suppose that 8% of the circuit boards in a newly shipped lot are defective. Using the sampling plan described above, calculate the probability that this lot will be accepted by the appliance manufacturer. Round your answer to at least three decimal places.

Thank you for any help. Believe me, I am not just pasting these on this site, I am looking at them for what seems like an eternity w/no clue how to redo them again.

My promise: I WILL NEVER ENTER THE KINGDOM OF STATS AGAIN WHEN I AM THROUGH WITH THIS COURSE.

2. Hello, Y-rag!

A manufacturer of electronic appliances buys circuit boards from an outside vendor.
The boards are shipped in large lots, with several thousand circuit boards per lot.
When each lot is received, a random sample of 30 boards is selected
and tested to see if the circuitry performs correctly.

If more than 5 of the circuit boards in the sample fail the testing,
then the entire lot is rejected and is shipped back to the vendor.

Suppose that 8% of the circuit boards in a newly shipped lot are defective.
Using the sampling plan described above, calculate the probability
that this lot will be accepted by the appliance manufacturer.
For Accept, there must be: 0, 1, 2, 3, 4, or 5 defective boards in the lot.

We must determine the probability of each case . . .

. . $\begin{array}{ccc}
P(\text{0 df}) &=& {30\choose0}(0.08)^0(0.92)^{30} \\ \\[-3mm]
P(\text{1 df}) &=& {30\choose1}(0.08)^1(0.92)^{29} \\ \\[-3mm]
P(\text{2 df}) &=& {30\choose2}(0.08)^2(0.92)^{28} \end{array}$
. . . $\begin{array}{ccc}
P(\text{3 df}) &=& {30\choose3}(0.08)^3(0.92)^{27} \\ \\[-3mm]
P(\text{4 df}) &=& {30\choose4}(0.08)^4(0.92)^{26} \\ \\[-3mm]
P(\text{5 df}) &=& {30\choose5}(0.08)^5(0.92)^{25}\end{array}$

. . . and add them.

3. Thank you Soroban. I was able to finally use the probability formula to figure it out. I will try your way to confirm. Thank you very much for the vest.

4. Soroban, the problem I gave you changed. The numbers were 46, 3 and 7% instead. That's what happens if I wait too long. My answer was 0.596 using the formula they gave. I believe that is right. I appreciate your help.

5. Originally Posted by Y-rag
Soroban, the problem I gave you changed. The numbers were 46, 3 and 7% instead. That's what happens if I wait too long. My answer was 0.596 using the formula they gave. I believe that is right. I appreciate your help.
Yes that is correct.
$\sum\limits_{k = 0}^3 {\binom{46}{k}\left( {.07} \right)^k \left( {.93} \right)^{46 - k} } =0.596354834867575$