Question

Just want to know if my procedure for this Binomial problem is correct:

Binomial problem:
Say \(Y_1,\ldots, Y_5\) be mutually independent and identical random variables with distribution Binomial(2, 0.3). Thus, each \(Y_i\) takes on values {0, 1, 2}. Define \(Z\) to be the number of \(Y_i\) which take on value "1". Example: \(\{Y_1,\ldots, Y_5\}= \{0,2,1,1,1\}\) would give \(Z=3\). Find \(\Pr(Z>4)\)

Answer 1

What I did: \(\Pr(Z > 4)=\Pr(Z=5)\) since Z can't exceed 5. Then I define an indicator random variable \(X_i\) such that

Answer 2

\[X_i=\begin{cases}
1 & \text{if } Y_i=1, prob={2\choose 1}(0.3)^1(0.7)^{2-1}=0.42\\
0 & \text{otherwise}\end{cases}\]

Answer 3

Then, \[Z=\sum_{i=1}^{5}X_i\sim \text{Binomial}(5, 0.42)\]So, \[P(Z=5)={5\choose 5}(0.42)^5(0.58)^{5-5}\]

Answer 4

Considering Y1-Y5 are IID, I would consider the experiment as a 5 step experiment, each with probability p (obtained from the binomial distribution).

Answer 5

@mathmate do you think you could elaborate a bit what you mean?

Answer 6

IID means "mutually independent and identical random variables".
You can calculate the probability of Yi taking the value of 1 using the binomial theorem, with \(r1=P(Z_1=1)=(^2_1)0.3^1(1-0.3)^1\)
Since r1=r2=r3=r4=r5 are all equal, and we need P(Z>4)=P(Z=5), this means that all trials have to give a 1. Use the multiplication law to give P(Z=5)=r1*r2*r3*r4*r5=r^5.

Alternatively, you can use the binomial theorem a second time to give
\(P(Z=5)=(^5_5) r^5(1-r)^0\) which gives the same result in this case.

If you had to calculate P(Z=4) or P(Z=3), definitely you will need to apply the binomial theorem a second time.

Answer 7

Thank you :) !

Answer 8

You're welcome! :)