MHB Digital Clocks: Binomial Problem Analysis and Results

  • Thread starter Thread starter Math101_McF
  • Start date Start date
  • Tags Tags
    Binomial
Math101_McF
Messages
2
Reaction score
0
A company makes digital clocks. It is determined that 5% of all clocks produced are defective.

you go to the warehouse and randomly select 80 clocks.
1. How many of the 80 clocks do you expect to be defective?

2.What is the probability that exactly 6 of the clocks are defective?

3. What is the probability that at least one of the clocks (out of 80) is defective? use the complement.
 
Physics news on Phys.org
Hi there,

Welcome to MHB! :)

We like to tackle one problem at a time. Which one do you want to look at? What are your thoughts on it?
 
1
 
Math101_McF said:
1
Ok so we really want to help you but won’t give you answers. If you show what you’ve done we will do a ton to get you to the finish line but if you want answers for free this isn’t the place.

What we do have here are volunteers with PhD’s, other advanced degrees, and years of experience teaching math. We actually want you to like math and learn. For free. Promise.
 
The answer is almost given in the question:
"A company makes digital clocks. It is determined that 5% of all clocks produced are defective.

you go to the warehouse and randomly select 80 clocks.
1. How many of the 80 clocks do you expect to be defective?"
5% of 80 is (0.05)(80)= 4.

"2. What is the probability exactly 6 clocks are defective."
Each clock is either "defective" or "not defective" so this is a "binomial distribution". There are 80 clocks. The probability any given clock is broken is 0.05 and the probability it isn't is 0.95. The probability exactly 6 out of 80 are broken is $\begin{pmatrix}80 \\ 6\end{pmatrix}(0.05)^6(0.95)^{74}$.

"3. What is the probability that at least one of the clocks (out of 80) is defective? use the complement."
The opposite of "at least one" is "none". Calculate the probability that none of the 80 clocks is defective, $(0.95)^{80}$ and subtract that from 1.
 
Last edited:
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...
Thread 'Detail of Diagonalization Lemma'
The following is more or less taken from page 6 of C. Smorynski's "Self-Reference and Modal Logic". (Springer, 1985) (I couldn't get raised brackets to indicate codification (Gödel numbering), so I use a box. The overline is assigning a name. The detail I would like clarification on is in the second step in the last line, where we have an m-overlined, and we substitute the expression for m. Are we saying that the name of a coded term is the same as the coded term? Thanks in advance.
Back
Top