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If X ~ Bern(p) and Y=nX is Y~Bin(n,p)?

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No. The sum of n Bernoulli trials would be a binomial.
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Nope. The binomial is saying to take n trials of biased coin flips, each with a probability p of heads independently, and count the number of them total.

Whereas, a Bern(p) is a single coin flip. nX is not the same thing, because it's n times the single flip, compared to doing n different flips.
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No. If X is a Bernoulli trial with probability of success p, that means X takes the value of 1 with probability p, and 0 with probability 1-p. Y = nX is also a similar to a Bernoulli trial, except is does not take the value of 1, it takes the value n with probability p.

Now, if Y = X1 + … + Xn, where X1, …, Xn are iid Bernoulli trials, then Y would be binomial(n, p).
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No.

One way to see this is

nX can only be two values: 0 or n

Bin(n, p) can be any number 0,1,2,3,...,n
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i think so
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