Probability of A and B and C occurring exactly once

Question

A spinner has 6 equal sectors; A,B,C,D,E,F. All have equal probability of 1/6. Find the probability of A and B and C exactly once if the spinner is spined 10 times. Also similarly find probability of getting A or B or C after spinning the spinner 25 times.

I did the enter image description here

I treated these as Bernoulli trials, probability of A exactly once comes to be 0.32 (rounded to 2 decimal), it would be same for B and C. I got the answer to first question as 0.57, but its wrong. This question is from a Python Hackathon online.

Answer 1

Part 1

Find the probability of A and B and C exactly once if the spinner is spined 10 times.

This doesn't look like a good candidate Bernoulli trials because the definition of success/failure doesn't remain consistent throughout the 10 trials. In one trial you want A, B in some other and C in a third. These are different outcomes, even if they happen to have the same probability.

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Here's one approach to solve this part.

We need A, B, and C each to appear exactly once. So, we are very particular about three spins - any three.

We can choose $3$ out of $10$ trials in $\left ( ^{10} _{\,3} \right) = 120$ ways.

The probability of getting A, then B, then C in one set of 3 trials would be

$$\left( \frac{1}{6} \right)^3$$

But we don't care in what order A, B, and C will appear. So all permutations of A, B, and C are good. So the total probability of getting A, B, and C (in any order) would be

$$ \begin{align*} 3! \times \left( \frac{1}{6} \right)^3 = \frac{1}{36} \end{align*}$$

Now the probability of not getting A, B , or C in one spin is the same as getting D, E, or F. And that would be

$$3 \times \frac{1}{6} = \frac{1}{2}$$

This must happen in all the seven remaining spins (leaving out the three reserved earlier). The probability would be

$$\left( \frac{1}{2} \right)^7 = \frac{1}{128}$$

So for one particular set of three spins (selected earlier), the probability of A, B, and C each appearing only once is

$$\frac{1}{36} \times \frac{1}{128}$$

Now, this is for only one set of three spins but it could be any of the $120$ possible combinations mentioned earlier. So total required probability is

$$ \begin{align*} 120 \times \frac{1}{36} \times \frac{1}{128} \approx 0.026 \end{align*}$$

Part 2

Also similarly find probability of getting A or B or C after spinning the spinner 25 times.

So the point is we must not get A, B or C in any of the first $25$ trials. That's all.

Here success is getting A or B or C. Every other outcome is a failure. The definition remains consistent and the two outcomes are mutually exclusive and exhaustive. We can go ahead with Bernoulli trials.

Also, probability of success is the same as that of failure, $1/2$.

Now, the probability of having $0$ successful trials in $25$ would be

$$ \begin{align*} P(X = 0) &= \, ^{25}C_{0} \times \left( \frac{1}{2}\right)^0 \times \left( \frac{1}{2}\right)^{25} \\[0.3cm] &= \frac{1}{2^{25}} \end{align*}$$

Answer 2

I understand. I don't want the answer. Just want to understand how to approach the problem. Can I treat this question as Bernoulli trials ?? – [@vikramchaudhari]

Somewhat.  You have 10 independent trials each with 4 possible states ($A, B, C$, or other) with constant rates of occurrence ($1/6, 1/6, 1/6$, and $3/6$).  This is analogous to a Binomial Distribution, but with multiple states rather than just two.  Thus, this kind of distribution is called a Multinomial Distribution.

A Binomial Distribution for $X$ successes among $n$ independent trials with identical success rate of $p$ (thus $n-X$ failures with failure rate $1-p$) has probability mass function: $$\mathsf P(X=x) =\binom n x p^x (1-p)^{n-x}\,\mathbf 1_{x\in[[0,...,n]]}$$

Likewise a Trinomial Distribution for $X$ state-1, $Y$ state-2, and $n-X-Y$ state-3 outcomes among $n$ independent 3-state trials with constate state-rates of $p, q, 1-p-q$ respectively, has joint probability mass function of : $$\begin{align}\mathsf P(X=x, Y=y) &= \binom{n}{x}\binom{n-x}{n-x-y} p^x q^y (1-p-q)^{n-x-y}\,\mathbf 1_{x\in[[0,..,n]], y\in[[0,..,n-x]]}\\[2ex] &=\dfrac{n!}{x!~y!~(n-x-y)!}\, p^x q^y (1-p-q)^{n-x-y}\,\mathbf 1_{x\in[[0,..,n]], y\in[[0,..,n-x]]}\end{align}$$

So, if you have $n$ independent trials ($10$ or $25$), with those $4$ states having the given occurrence rates ....

Answer 3

Question 1...

Think in terms of sample spaces, each equally probable outcome being a sequence of length 10 consisting of the letters A to F. There are $6^{10}$ of these.

To count "successful" outcomes there are $3!\binom{10}3$ of placing the letters A, B, C and then $3^7$ ways of filling the remaining 7 spaces with the other 3 letters.

Question 2...

Here the sample space has size $6^{25}$ elements, "successful" elements are all of these except from the $3^{25}$ elements that only use the letters D, E, F