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# Advanced Math/Possibilities of drawing 3 numbers with 5 tries.

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QUESTION: I am trying to calculate the possibilities of having the following scenario:

Given a range of balls with number from 1 to 20 (1,2,3,4,5...20), what is the possibilities of me picking up ball number 1,2,3 (in any sequence) in 5 tries?

Thank you very much for your time on helping this question.

ANSWER: Eugene, The question is a little too ambiguous so let's try to sort it out. Do you want to know the probability of ending up with the balls 1, 2, 3, in any order, after picking up one ball in 5 tries? That is, out of the 5 balls picked in the 5 tries (without replacement), you end up with the 1, 2, 3 balls in the mix? Please clarify.

---------- FOLLOW-UP ----------

QUESTION: Hi Randy:

Yes. I would like to know the probability of ending up with balls of 1,2,3 (in any order) after picking one ball at a time with 5 tries. Ball that are picked will not be replaced (Hence first try will be 1/20, second will be 1/19, third 1/18...).

Of the 5 balls picked, what is the probability of me picking up ball of 1,2 and 3 in the mix.

I hope I have made the question clearer.

Thank you very much in advance.

ANSWER: Eugene, Here's how I figured it.

Out of 20 well shuffled numbered balls, what is the probability of choosing the numbers 1,2 3 in any order if the balls are sampled 5 times without replacement?

To start, there is more than one way to choose the 1,2,3 balls in the 5 tries. The number of ways is given by the number of permutations of 3 balls out of 5, since different orders are counted separately, or

N = 5!/2! = 60.

An example of the sequence of choosing balls is (with the balls now labeled A,B,C for clarity)

A B # C #    where # denotes any ball not 1,2, or 3. Other examples are
A # B # C
# # A B C
etc.

Note that, for instance

# # A B C   and
# # C B A

are considered separate permutations.

Now we need to calculate the probability of each of the 60 combinations. To do this, we can use the property of probabilities

P(a∩b∩c∩d∩e) = P(a)･P(b|a)･P(c|a,b)･P(d|a,b,c)･P(e|a,b,c,d)

where a,b,c,d,e are the events of picking A,B,C or #, a∩b means both events a and b occur and the notation P(b|a) = probability of event b given that event a has occurred.

Let's first take the sequence where N = A,B or C

N N N # #.

The probabilities are (as the balls are sampled without replacement)

P(N) = 3/20 since any of A,B or C could have been chosen
P(N|N) =(3-1)/19 =  2/19 since one of A,B,C has been chosen, leaving 2 left, and the sample size is now 19.
P(N|N,N) = (3-2)/18 = 1/18  for simialr reasoning
P(#|N,N,N) = (20-3)/17 = 17/17
P(#|N,N,N,#) = (20-4)/16 = 16/16

so that

P(N∩N∩N∩#∩#) = (3･2･1･17･16)/(20･19･18･17･16) = (3･2･1)/(20･19･18) ≈ 8.8x10^-4.

If the other sequences are evaluated in the same way, for instance P(N,#,N,#,N), it can be seen that the probabilites are equal for each of the 60 sequences. Therefore, the total probability of getting 1,2,3 is (9.9^10-4)･(60) ≈ 0.053.

I think!

---------- FOLLOW-UP ----------

QUESTION: Hi Randy, I did a computer generated simulation and the figure is very close to your calculation.

I would like to further this question with a different scenerio.

What is the possibility of drawing two balls of any number 1, 2 or 3 of number 1,2,3,4,5 from the 5 picks out from the pool of 20 balls with numbers 1,2,3,4...20?

In this case, drawing of the following number is considered valid:

1,2,#,#,#
1,3,#,#,#
3,2,#,#,#
#,#,1,2,#
#,#,1,#,3

but not
1,2,3,#,#
1,2,#,3,#
#,1,2,3,#

Specifically, I want to know only 2 balls hitting the right number.

The previous question was 1,2 AND 3. How do we calculate for OR scenerio?

Thank you in advance.

The way I solve this is to calculate the probability of getting 2 of the specific balls, in a manner similar to the previous problem, and then calculating how many ways this combination can come up out of the 3 possible balls.

Using the notation and reasoning from before, the result from the 5 picks, for example, for the pair N = 1 & 2

N,N,#,#,#  or
#,N,#,#,N  or
N,#,N,#,#

...

have the probability (2/20)･(1/19)･(18/18)･(17/17)･(16/16) = 2/(20･19) = 5.3x10^-3 = 0.0053.

Now, there are 5!/(5-2)! = 5!/3! = 20 ways to arrange these 2 numbers in the result.

We also have 5!/(3!2!) = 3 ways to choose a pair from the numbers 1,2,3, so altogether, the probability is

(0.0053)･(20)･(3) = 0.318.

This seems a little high to me but there you have it.

I have used the property of probabilities that for independent events, the union E1∪E2 has the probability P(E1∪E2) = P(E1) + P(E2). The union means E1 or E2.

It is interesting to note that the probability of 3 numbers calculated as part of the original problem above could have been obtained by calculating the number of permutations of 20 numbers taken 3 at a time, i.e., 20!/(20-3)! = 6800, along with the number of ways of arranging 3 numbers, 3! = 6 to get

6/6800 = 8.8x10^-3.
Questioner's Rating
 Rating(1-10) Knowledgeability = 10 Clarity of Response = 10 Politeness = 10 Comment Thanks alot! I am going through this and is testing out it over and over again. I appreciate your help. Cheers!

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#### randy patton

##### Expertise

college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography

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26 years as a professional scientist conducting academic quality research on mostly classified projects involving math/physics modeling and simulation, data analysis and signal processing, instrument development; often ocean related

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J. Physical Oceanography, 1984 "A Numerical Model for Low-Frequency Equatorial Dynamics", with M. Cane

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M.S. MIT Physical Oceanography, B.S. UC Berkeley Applied Math

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