Question

3- (20 points) A random experiment consists of simultaneously and independently flipping a coin five times and observing the n-5 resulting values facing up. The coin is biased with: P(heads) - 0.75 : P(tails) p-0.25 Define a Random Variable (RV) X equal to the number of fails that we observe during the flips. a) Give the probability P. that the random variable X will take on the value 3 ANSWER: P,= (simplified number) b) Give the mean of X, that is, give the value of E[X]. ANSWER: E[X] - (simplified number) c) Use a Poisson random variable Y to model this random experiment in the best possible way that matches a Binomial RV. Give the probability P. that the random variable Y will take on the value 3 and compare this to the answer in part a). Also, indicate which one is bigger P. or P.?

Answer 1

Here $X \sim Binomial(5,0.25)$

(a) We are to calculate P_a=P(X=3)

$P(X=3)=\binom{5}{3}(0.25)^{3}(1-0.25)^{5-3}=\binom{5}{3}(0.25)^{3}(0.75)^{2}=0.087890625$

(b) The mean of X is given by

$E(X)=\sum_{i=0}^{5}(X_{i}*P(X_{i}))$

By the property of binomial random variable, we know that the mean oof the distribution is the product of the parameters, if

$Z \sim Binomial(n,p)$ then the mean of the distribution of Z is np.

Here the mean is then given by

$E(X)=5*0.25=1.25$

(c) Though Poisson random variable cannot model the distribution but considering Y as the number of tails obtained in 5 throws of a fair coin.

Hence by the law of approximation of binomial random variable to Poisson random variable, we can say

$Y \sim Poisson(5*0.25)$

$i.e. \;Y \sim Poisson(1.25)$

Then Pc=P(Y=3) is given by

$P(Y=3)=\frac{e^{-1.25}*(1.25)^{3}}{3!}=0.09326328023$

Now we have obtained P_a= 0.087890625 and P_c=0.09326328023

i.e P_c>P_a