Probability of Selecting a Black Disc

[karush]

A box contains 35 red discs and 5 black discs
a disc is selected at random and its color noted.
The disc is then replaced in the box.

a) In 8 such selections what is the probability that a black disk is selected.
i) exactly once ii) at least once

b) The process of selecting and replacing is carried out 400 times
what is the expected number of black discs that would be drawn.

well there are 40 discs so the probability of selecting a black disk it 1 to 8 so i would presume "at least once" is 8 times. but not sure about the "exactly once"

about 400 times, not sure how this is done I just 400/8 to get 50

I don't know the answers to these so just see what would be suggested here.

 

[MarkFL]

Re: disc selection

For part a), I would look at the binomial probability formula:

\(\displaystyle P(x)={n \choose x}p^x(1-n)^{n-x}\)

The binomial coefficient \(\displaystyle {n \choose x}\) takes into account the number of ways to get $x$ successes for $n$ trials.

Here $p$ is the probability of drawing a black disc on one trial, $n$ is the number of trials, and $x$ is the number of successes.

For part ii), I would use the fact that it is certain we will either draw zero black discs OR we will draw at least one black disc. This will make your computation much simpler.

Can you show what you can do with these suggestions?

 

[HallsofIvy]

A box contains 35 red discs and 5 black discs
a disc is selected at random and its color noted.
The disc is then replaced in the box.

a) In 8 such selections what is the probability that a black disk is selected.
i) exactly once ii) at least once

b) The process of selecting and replacing is carried out 400 times
what is the expected number of black discs that would be drawn.

well there are 40 discs so the probability of selecting a black disk it 1 to 8 so i would presume "at least once" is 8 times.

I'm afraid you are completely misunderstanding the question. A probability is a number between 0 and 1 so "8 times" is impossible.

but not sure about the "exactly once"

about 400 times, not sure how this is done I just 400/8 to get 50

As the answer to what question?

I don't know the answers to these so just see what would be suggested here.

 

[karush]

Re: disc selection

For part a), I would look at the binomial probability formula:

\(\displaystyle P(x)={n \choose x}p^x(1-n)^{n-x}\)

The binomial coefficient \(\displaystyle {n \choose x}\) takes into account the number of ways to get $x$ successes for $n$ trials.

Here $p$ is the probability of drawing a black disc on one trial, $n$ is the number of trials, and $x$ is the number of successes.

For part ii), I would use the fact that it is certain we will either draw zero black discs OR we will draw at least one black disc. This will make your computation much simpler.

Can you show what you can do with these suggestions?

sorry so long to get back to this, but we went on to another topic next day, but still want to deal with this

from \(\displaystyle {n \choose x} \) just to get the coefficient i presume \(\displaystyle n = 8\) and \(\displaystyle x = 1\) since the ratio is 5:40 or 1:8 so the coefficient would be 8

 

[MarkFL]

Re: disc selection

sorry so long to get back to this, but we went on to another topic next day, but still want to deal with this

from \(\displaystyle {n \choose x} \) just to get the coefficient i presume \(\displaystyle n = 8\) and \(\displaystyle x = 1\) since the ratio is 5:40 or 1:8 so the coefficient would be 8

Yes, there are 8 ways to get exactly in black disk, one for each drawing of a disk from the box.