# Binomial problem.

#### Math101_McF

##### New member
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.

#### Jameson

Staff member
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?

#### Math101_McF

##### New member
1

• topsquark and Jameson

#### Jameson

Staff member
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.

• topsquark and MarkFL

#### Country Boy

##### Well-known member
MHB Math Helper
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.

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