Probability with a twist

Hello, thanks for taking the time to read the following query. Lets say there are 13 balls in a bag numbered 1-13. Before any balls are drawn I select 7 numbers and then 7 balls are drawn from the bag. I can work out the chances of all 7 selections being drawn (13choose7) . I can also work out the chances that say 3 of my 7 selections are drawn (7choose3*6choose4/13choose7). I CAN'T, however, work out the probability for the following scenario. Let's say again there are 13 numbered balls and 7 to be drawn out. What is the probabilty that if I only select 5 numbers initially before 7 are drawn, 3 of my selections are drawn? I would like to be shown how to work this out using binomials without having to list all the combinations which took me ages! thank you

Hello, goodminton!

There are 13 balls in a bag, numbered 1 to 13.
Before any balls are drawn I select 7 numbers.
Then 7 balls are drawn from the bag.
What is the probability that 3 of my numbers are chosen?

\(\displaystyle \text{I can work out the number of possible outcomes: }\:_{13}C_7\)
\(\displaystyle \text{I can also work out the probability that 3 of my 7 selections are drawn: }\:\frac{(_7C_3)(_6C_4)}{_{13}C_7}\)

I CAN'T, however, work out the probability for the following scenario.
Let's say again there are 13 numbered balls and 7 to be drawn.
If I select 5 numbers initially, what is the probability that 3 of my selections are drawn?

I would like to be shown how to work this out using binomials
without having to list all the combinations which took me ages! . . . . What?

Click to expand...

You're kidding, right?

\(\displaystyle \text{There are: }\:_{13}C_7 \:=\:\frac{13!}{7!\,6!}\text{ possible outcomes.}\)

I doubt that you listed all 1,716 outcomes and picked out the ones you wanted.
Besides, if you did that, why did you write out all those combination numbers?

\(\displaystyle \text{Your first answer is correct: }\:\frac{(_7C_3)(_6_4)}{_{13}C_7} \;=\;\frac{525}{1716}\)

\(\displaystyle \text{The second answer is: }\;\frac{(_5C_3)(_8C_4)}{_{13}C_7} \;=\;\frac{700}{1716} \;=\;\frac{175}{429}\)

Get the idea?

goodminton said:

Hello, thanks for taking the time to read the following query. Lets say there are 13 balls in a bag numbered 1-13. Before any balls are drawn I select 7 numbers and then 7 balls are drawn from the bag. I can work out the chances of all 7 selections being drawn (13choose7) . I can also work out the chances that say 3 of my 7 selections are drawn (7choose3*6choose4/13choose7). I CAN'T, however, work out the probability for the following scenario. Let's say again there are 13 numbered balls and 7 to be drawn out. What is the probabilty that if I only select 5 numbers initially before 7 are drawn, 3 of my selections are drawn? I would like to be shown how to work this out using binomials without having to list all the combinations which took me ages! thank you

Click to expand...

You have worked out that exactly three of your selections match.
But there could be three or more.
\(\displaystyle \sum\limits_{k = 3}^7 {\binom{7}{k}\binom{6}{7-k}}\)

Ya, that's one with a twist for sure.

Soroban and Pka; his problem needs not contain that introduction about what he can do, right?
This is all that's required verbiositically(!) typing:
A bag contains 13 balls numbered 1-13 and 7 are drawn out at random.
What is the probabilty that if I select 5 numbers before the 7 are drawn, 3 of my selections are drawn?
That's his problem in full, right?

And it can be worded this way:
A bag contains 13 balls numbered 1-13 and 7 are drawn out at random.
Now I select 5 numbers. What is the probability that 3 of my selections have been drawn?
Same thing, right?

So there are 7 numbers drawn which I can't see, and I write down 5 numbers; and hope to
write 3 OR MORE that are in the 7 drawn.

Well, I ain't versatile enough to work that out "nicely", but I'm giving 2 to 1 odds that the
probability is over 50%. Much higher than Soroban's (so I'm nervous!).

Actually, I wrote a simulation program that indicates ~58.75 %

Thank you Soroban and pka for the help. Soroban, I did write out combinations but that was the 56 from 8balls when 5 are drawn. The reason i listed these was to check for sure the probability of having 3 numbers in the 5 drawn when initially selecting 4. I did want to apply this to more complex problems and you have helped me to do this. pka, the 'sum of' advice has also helped me. Thank you also Denis for the English lesson