MHB Probability of Getting 3 A's & 2 B's in 5 Rounds

  • Thread starter Thread starter oriel1
  • Start date Start date
AI Thread Summary
The discussion focuses on calculating the probability of obtaining three A's and two B's from a machine with three possible outcomes: A, B, and C. The initial calculation using the formula for combinations and probabilities was deemed correct, despite concerns about the result being greater than 1. It was clarified that while the binomial distribution is not applicable due to the presence of three outcomes, the multinomial distribution can be used instead. The final probability calculated is small, which some participants found surprising. The conversation emphasizes the importance of understanding the appropriate statistical methods for different scenarios.
oriel1
Messages
8
Reaction score
0
There is lucky machine with 3 results (letters):
A,B,C
P(A)=0.2 (Probabilty to get A)
P(B)=0.3 "" "" ""
P(C)=0.5
Each round you get just one letter.
You played 5 times.
What the probability that you got three times "A" and two times "B".

i thinked to do that:
$$0.2^{3}$$$$\cdot*$$$$0.3^{2}$$$$\frac{{5}!}{{2}!\left({3}\right)!}$$
$$\frac{{5}!}{{2}!\left({3}\right)!}$$ is the number of options to get that result.
But it doesn't make any sense. because $$\frac{{5}!}{{2}!\left({3}\right)!}$$=10 and if the probabilty is higher then 0.1 the result will be >1 and it can't happen.
There is any good solution for this problem ? binomial distribution or something?
Thank you.
 
Mathematics news on Phys.org
Hi oriel,

Your calculation looks good to me! You found the probability of one sequence of 3 A's and 2 B's, then multiplied by the number of sequences. Indeed if $(.2^3 \cdot .3^2)$ were greater than .1, then we would have a problem, but it isn't. If you multiply your answer out, we get a valid probability. This isn't proof that the answer is correct, but it isn't outside the range of a probability calculation.
 
I was typing an answer but then I saw Jameson's post and he basically gave the answer. Your calculation is fine! Now, I want to add a comment to your final question: can we use a binomial distribution here? The answer is no. The binomial distribution can only be used in experiments where you have two possible outcomes (a failure and a success). Here you have three. However, you can use the multinomial distribution which is a generalization of the binomial distribution for $k$ different outcomes. If you're interested then I suggest you read the wikipedia page:
https://en.wikipedia.org/wiki/Multinomial_distribution

You'll notice that your calculation is exactly how the probability would be calculated if you had explicitly used the multinomial distribution.

Finally, the probability is indeed small ... do you find the result surprising?
 
Last edited:
Jameson said:
Hi oriel,

Your calculation looks good to me! You found the probability of one sequence of 3 A's and 2 B's, then multiplied by the number of sequences. Indeed if $(.2^3 \cdot .3^2)$ were greater than .1, then we would have a problem, but it isn't. If you multiply your answer out, we get a valid probability. This isn't proof that the answer is correct, but it isn't outside the range of a probability calculation.
Siron said:
I was typing an answer but then I saw Jameson's post and he basically gave the answer. Your calculation is fine! Now, I want to add a comment to your final question: can we use a binomial distribution here? The answer is no. The binomial distribution can only be used in experiments where you have two possible outcomes (a failure and a success). Here you have three. However, you can use the multinomial distribution which is a generalization of the binomial distribution for $k$ different outcomes. If you're interested then I suggest you read the wikipedia page:
https://en.wikipedia.org/wiki/Multinomial_distribution

You'll notice that your calculation is exactly how the probability would be calculated if you had explicitly used the multinomial distribution.

Finally, the probability is indeed small ... do you find the result surprising?
Thank you very much for that clear explanation. I appreciate that.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top