A bag containing 1 green, one red and 5 white balls.

I withdraw a ball and then without replacing it withdraw another. What is the probability that I withdraw both red and green balls?

So, I know:
Red = 1/7
Green = 1/7
White 4/7 (?)

For the sake of the equation, do I ignore the white balls, thus give it as:

1/7 * 1/7 = 1/49 or would the 2 chances merge to make 2/49?

 

Calculate the total number of possible two-ball combinations - 21. 1/21 then.

 

The chance of drawing a red on the first try is 1/7, yes?

Also, the chance of drawing green on the first try is 1/7, yes?

So, the chance of drawing red -or- green on the first try is what?

Then, 6 balls remain, with either a single red or a single green.

What are the odds of drawing the remaining colored ball?

Combine the first and second answer (i.e. multiply).

 

i think you're confused and you made me confused as well with your last comment so i'll just do the math straight forward with explanations. Here we go;

If i get you right, we have 2 desired situations here. I'll handle them separately.
Situation 1;
You withdraw a red ball first with the probability of 1/7. Then you repeat it without dropping it back. Now you withdraw a green ball with the probability of 1/6. So, for situation 1, your desired probability is (1/7)*(1/6)=1/42
Situation 2;
You withdraw a green ball first with the probability of 1/7. Then you repeat it without dropping it back. Now you withdraw a red ball with the probability of 1/6. So, for situation 2, your desired probability is (1/7)*(1/6)=1/42

The last step is summing these probabilities up, since there is a 'or' relation between situation 1 and situation 2;
(1/42)+(1/42)=1/21

Hope this helps. If you need any further explanation, feel free to ask.

 

Calculate the total number of possible two-ball combinations - 21. 1/21 then.

What math did you use to get 21? Just curious. I used Aerb's solution which I thought was the quickest way to solve.

 

This is 'one of them' ones that I understand the principle but have a git of a time trying to put on paper - Thanks.

So I have 2 more senarios;

1 -I replace the balls, shake the bag and then withdraw 2 balls together. What is the probability that I have with drawn both the red and green balls?


2 - All the balls are returned to the bag and 4 are removed together. What si the probability that both red and green balls are left in the bag?

For both of these, is it a simple case of using 2 and respectively instead of 1?

 

What math did you use to get 21? Just curious. I used Aerb's solution which I thought was the quickest way to solve.

Use the "Binomial Coefficient" to calculate the number of combinations of r objects of n items, where r=2 and n=7.

\(\frac{7!}{2!(7-2)!}\) = \(\frac{5040}{2*120}\)

This is equal to 21!

 

For your latest scenarios, the first is identical to the original problem, since no matter what, the second ball (even if picked at the same time) is still coming from a set of 6 elements.

The second scenario can be thought of as the inverse of the first.
Hence if \(P_{1}\) is equal to the probability of the first scenario, then \(1-P_{1}\) is the probability of the second.

 

What math did you use to get 21?.

That's the formula I used. But it's simpler than it seems because n!/(n-r)! is quite easy to calculate. If r is 1, it's n. If r is 2, it's n*(n-1). If r is 3, it's n*(n-1)*(n-2) etc.

 

2 - All the balls are returned to the bag and 4 are removed together. What si the probability that both red and green balls are left in the bag?

What is the number of possible combinations of 4 balls out of 7 - 35.

How many meet our criteria? It is equal to the number of combinations of 4 out of 5 white balls - 5.

Probability: 5/35 = 1/7.

Just to make sure you didn't make any mistakes: you'll get 10 situations where you get a green ball with all others white, 10 situations where you get a red ball and all others white, and 10 situations where you get green, red and two whites.

10+10+10+5=35.

Do they make you draw the real balls many many times and verify that the result converges to expected probabilities?

 

For your latest scenarios, the first is identical to the original problem, since no matter what, the second ball (even if picked at the same time) is still coming from a set of 6 elements.

Just to confirm, you mean this: All the balls are returned to the bag and 4 are removed together. What is the probability that both the red and green balls are left in he bag?

- This you say is the same method as the first?

 

What is the number of possible combinations of 4 balls out of 7 - 35.

How many meet our criteria? It is equal to the number of combinations of 4 out of 5 white balls - 5.

Probability: 5/35 = 1/7.

Please can you clarify how 5/35 = 1/7 is there an equation ?

 

Please can you clarify how 5/35 = 1/7 is there an equation ?

No, there's a table. You had to memorise it at school: 5 x 7 = 35. I don't know if they still teach it ...

 

No, there's a table. You had to memorise it at school: 5 x 7 = 35. I don't know if they still teach it ...

I was going to reply in a similar fashion, but it just seemed harsh. Even for me.

 

Us Brits say you North Americans have no sense of humour.....

Regardless, coming from educated folk such as yourself still **** poor. My Bad....

 

Us Brits say you North Americans have no sense of humour.....

Yes, English sense of humor in times of Charles Dickens was so good that some Brits still believe they have it

 

Ha...which is it English or British? Do you know the difference?

 

Ha...which is it English or British?

Who? Famous sense of humor? Definitely English. Dickens? Both English and British at the same time. Brits? Some are still English ...

 

with regards to the 10+10+10+5, is 5 representive of ???

 

with regards to the 10+10+10+5, is 5 representive of ???

It'll be more if Scottland separates.

Real-life application of probabilities