We will now look at the binomial in general. (a) [harder] Show using the denitio

Question

We will now look at the binomial in general.

(a) [harder] Show using the denition of equals in distribution that X1 d= X2 if X1 Bernoulli (p) and X2 Binomial (1; p).

(b) [harder] Imagine an innite bag where 47% of the balls are successes. If I draw 87 balls, what is the probability I get 29 success balls?

c) [harder] Imagine I have a bag with 300 balls where 141 of the balls are successes. If I draw 87 balls with replacement, what is the probability I get 29 success balls?

(d) [harder] Why is your answers to (c) and (d) the same?

(e) [easy] Let X1; : : : ;Xn iid Bernoulli (p). Give a real-life example of this situation 10

(f) [easy] Let Tn = X1+: : :+Xn where X1; : : : ;Xn ~iid Bernoulli (p). How is Tn distributed?

Explanation / Answer

b) Imagine an innite bag where 47% of the balls are successes. If I draw 87 balls, what is the probability I get 29 success balls?

Here number of successes follows Binomial distribution with parameters n = 87 and p = 47% = 0.47

And we have to find P(X = 29)

P(X=29) = (87 C 29) * 0.4729 * (1 - 0.47)(87-29) = 0.0032

(e) [easy] Let X1; : : : ;Xn iid Bernoulli (p). Give a real-life example of this situation 10

Tossing a coin.

It has two outcomes head and tail which has probability 1/2 for each outcome.

(f) [easy] Let Tn = X1+: : :+Xn where X1; : : : ;Xn ~iid Bernoulli (p). How is Tn distributed?

Tn follows Binomial (n,p)

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