cumulative probability of an event happening

Question

A production line is making 4 color (red, blue,green,yellow) gumballs and dropping 7 balls in a small bottle and 24 balls in a large bottle. Both bottles are sold as a pack. All balls have equal probability and the production line is making a 10 thousand a day.

What is the probability that the smaller bottle has exactly one red ball, and the larger bottle has exactly 7 red balls.

Will the probability of smaller bottle be multiplied with larger bottle. How to calculate the cumulative probability ? Any ideas anyone please.

Answer 1

Well, if this is it, there are some ambiguities, or at least some issues to make more precise.

We need to assume that:

• Every color has the same probability on each pick.
• Every pick is independent of the previous ones (which is reasonable if we have a large quantity of gumballs, much more than 7 and 24).
• Large and small bottles are filled in a way that the content of one is independent of the others (for instance, one is filled after the other; or there are different machines for each type; etc.).

Under those circumstances, if we call $R_S$ and $R_L$ to the number of red balls in the small and the large bottle, respectively, then:

• Yes, independence between bottles imply $$P(R_S=1 \wedge R_L=7)=P(R_S=1) \cdot P(R_L=7).$$
• Every time you pick a ball, the probability that it be red is $\tfrac14$.
• When you pick the balls for the small bottle you are repeating 7 times the same random experience, and everytime the probability that the ball be red is the same... so what do you think is the distribution of $R_S$?
• And what about $R_L$?

(The answer, at the end).

If you know that, then is easy to do the calculations since, as we said, we want $$P(R_S=1 \wedge R_L=7) = P(R_S=1) \cdot P(R_L=7).$$

But I don't quite get to which random variable you refer when you ask for the cumulative distribution. $R_S$? $R_L$? Their sum? Something else?

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ANSWER: It's clear that $$R_S \sim Bin(7,0.25) \quad \text{and} \quad R_L \sim Bin(24,0.25).$$

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UPDATE: That probability (the 1,7 combination) is the number you get in the previous calculations. Let's call that result $P(R_S=1 \wedge R_L=7):=p_*$.

Then, to discuss how many times will that haven among $n$ packs, has nothing to do with cumulative probabilities of the variables we've been defining. Now you are repeating $n$ independent times (n=1000, 5000, etc.) an experiment (preparing a pack and checking if it has the desired number of red balls); so the number of "successful" packages (say $X$) among $n$ is again a binomial r.v., now $X \sim Bin(n,p_*)$.

You can answer now any question about $X$. In particular, if you want the expected number of "successful" packages, remember that since it is a binomial distribution you will have $E(X)=np_*$ (and $Var(X)=np_*(1-p_*)$, in case you're wondering). And also remember that for 'large' $n$ you can answer questions about $X$ using an approximation with the normal distribution.