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I'm solving the following homework task:

There are 2 bins: A and B, and the probability for a ball to be placed in the bin A is 30% higher than in the bin B. 15 balls are being independently distributed among the bins. What is the probability that exactly 10 balls are put in the same bin?

Here is my solution:

P(A) + P(B) = 1
P(A) - P(B) = 0.3

So, the probabilities are

P(A) = 0.65
P(B) = 0.35

Then

P{10 balls are put in the same bin} =
  P{10 balls are put in the bin A} + P{10 balls are put in the bin B}

Using the binomial distribution for a ball to be placed in a bin

X ~ Binomial(15, 0.65), Y ~ Binomial(15, 0.35)
P{X = 10} + P{Y = 10}

I get the following number (I use R for the calculations):

> dbinom(10, 15, 0.65) + dbinom(10, 15, 0.35)
[1] 0.2219504

Does this look correct?

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    $\begingroup$ Downvote is mine. Choose a better name for the question and I'll remove it :) $\endgroup$
    – Firebug
    Feb 11 '18 at 14:10
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Everything is correct.

Except that the phrase "the probability for a ball to be placed in the bin A is 30% higher than in the bin B" is ambiguous. It could mean either $P(A)-P(B)=0.3$ or $P(A)=1.3P(B)$.

Ideally, you should say a word about events $(X=10)$ and $(Y=10)$ being incompatible so that you can sum their probabilities (which is clearly the case).

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    $\begingroup$ +1 for the bit about ambiguity. Perhaps the OP should look for a course where they set sensible questions. $\endgroup$
    – mdewey
    Feb 11 '18 at 13:41

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