Balls in a jar

Muffins

New member
Joined
Aug 24, 2013
Messages
15
I wonder if my approach to this is correct-

Say I have a jar with 10 balls, 9 of which are green and 1 that is red. I start taking out a ball at a time until I get the red ball out. What's the probability of taking 6 balls out? (No repetition)

I thought this might be a case of conditional probability? Basically taking 6 balls out would mean the 6th ball has to be red. so 9/10*8/9*7/8*6/7*5/6*1/5=0.1

Is this correct?

Thanks
 
I wonder if my approach to this is correct-

Say I have a jar with 10 balls, 9 of which are green and 1 that is red. I start taking out a ball at a time until I get the red ball out. What's the probability of taking 6 balls out? (No repetition)

I thought this might be a case of conditional probability? Basically taking 6 balls out would mean the 6th ball has to be red. so 9/10*8/9*7/8*6/7*5/6*1/5=0.1

Is this correct?

Thanks
Yes, that is the solution for "exactly 6" balls drawn. :)

It would also be possible to interpret the question as being "at least 6" instead of "exactly 6." In that case you would only multiply the first five of your six probabilities.
 
Hello, Muffins!

Say I have a jar with 10 balls: 9 are green and 1 is red.
I start taking out a ball at a time until I get the red ball.
What's the probability of taking 6 balls out? (No repetition)

Basically taking 6 balls out would mean the 6th ball has to be red.

So: .\(\displaystyle \displaystyle\frac{9}{10}\cdot\frac{8}{9}\cdot \frac{7}{8}\cdot \frac{6}{7}\cdot\frac{5}{6}\cdot\frac{1}{5}\;=\; \frac{1}{10}\)

Is this correct?

Yes, that is correct!

Here's another approach.

Take out the balls one at a time and place them in a row.

. . \(\displaystyle \circ\,\circ\,\circ\,\circ\,\circ\,\circ\,\circ\, \circ\,\circ\,\circ \)


What is the probability that the red ball is the sixth ball?
. . \(\displaystyle P(\text{Red is 6th}) \:=\:\frac{1}{10}\)
 
Need help with part 2 of this question

So the next part got me a bit. I am more confused than I care to admit.

Same jar - 10 balls- 1 red, 9 green.
3 balls are taken out without replacement, several times. (we don't know how many times).
*Clarification: Every 3 balls we take out, we return them back into the jar again, mix the jar, take out 3 again and so on.
Now here comes the monstrous question:
What's the probability that the third time in which one of the balls is red, is the sixth time 3 balls were taken out?

I thought this might be a case of binomial distribution, so I'd call a random variable X to be 'the amount of times we took 3 balls out, one of which is red'
X~B(6,p)
I gathered that P is 0.3 because there are 3 ways/orders in which we can draw a red ball out of a group of 3 balls, and the probability to draw a red ball out of the 10 in the jar is 1/10. so 1/10*3 gives 0.3.
Then it's up to putting things in the formula:
P(x=3)= (3 out of 6)*(0.3)^3*(0.7)^3

However I think that if I pick binomial distribution - order doesn't matter, and here clearly I need the third time of the sixth.

Help!
 
So the next part got me a bit. I am more confused than I care to admit.

Same jar - 10 balls- 1 red, 9 green.
3 balls are taken out without replacement, several times. (we don't know how many times).
*Clarification: Every 3 balls we take out, we return them back into the jar again, mix the jar, take out 3 again and so on.
Now here comes the monstrous question:
What's the probability that the third time in which one of the balls is red, is the sixth time 3 balls were taken out?

I thought this might be a case of binomial distribution, so I'd call a random variable X to be 'the amount of times we took 3 balls out, one of which is red'
X~B(6,p)
I gathered that P is 0.3 because there are 3 ways/orders in which we can draw a red ball out of a group of 3 balls, and the probability to draw a red ball out of the 10 in the jar is 1/10. so 1/10*3 gives 0.3.
Then it's up to putting things in the formula:
P(x=3)= (3 out of 6)*(0.3)^3*(0.7)^3

However I think that if I pick binomial distribution - order doesn't matter, and here clearly I need the third time of the sixth.

Help!
From the way the problem was stated, I was also thinking the binomial distribution might be useful, But I came to the same conclusion as you, that the order of the tests matters, so we can't use the distribution in the usual way. You have

\(\displaystyle p = 0.3 \),......\(\displaystyle n = 6\),......\(\displaystyle x = 3\)

\(\displaystyle \displaystyle P(x=3) = 0.3^3\ 0.7^3\ \binom{6}{3} = 0.18522\)

To finish off you have to do a conditional probability. Given that \(\displaystyle n=6\) and \(\displaystyle x=3\), what is the probability that event number 6 is a "true" event? Multiply that conditional probability by \(\displaystyle P(x=3)\) to get the final result.
 
Hello, Muffins!

Same jar - 10 balls: 1 red, 9 green.
3 balls are taken out with replacement several times.

Now here comes the monstrous question:
What's the probability that the third time in which one of the balls is red
is the sixth time 3 balls were taken out?

Let the action of taking out 3 balls be called one "draw".

In a draw, the probability of getting the Red ball is:
. . \(\displaystyle {3\choose2}(\frac{9}{10})(\frac{8}{9})(\frac{1}{8})\:=\:\frac{3}{10}\)

The probability of not getting the Red ball is:
. . \(\displaystyle (\frac{9}{10})(\frac{8}{9})(\frac{7}{8}) \:=\:\frac{7}{10}\)


In the first 5 draws, we want to get the Red ball twice: .\(\displaystyle {5\choose2}(\frac{3}{10})^2(\frac{7}{10})^3 \)
In the 6th draw, we want to get the Red ball: .\(\displaystyle \frac{3}{10}\)


Therefore: .\(\displaystyle \dfrac{3087}{10,\!000}\cdot\dfrac{3}{10} \;=\;\dfrac{9261}{100,\!000}\)
 
Top