1. ## Binomial distribution

Hello!

Q: In a large population, the proportion having blood group A is 35%. Specimens of blood from the first five people attending a clinic are to be tested. It can be assumed that these five people are a random sample from the population. The random variable X denotes the number of people in the sample who are found to have blood group A.

(i) Show that $P (X \leq 2) = 0.765$, correct to 3 significant figures.

(ii) Four such samples of five people are taken. Find the probability that three of these samples each have more than two people with blood group A. [Answer is 0.0398 (to 3 sig. fig.)]

(iii) Find the probability that the first three of these four samples each has exactly two people with blood group A, given that the total number from these four samples having blood group A is 9.

===

I managed to solve parts (i) and (ii) but got stuck at (iii). I know what conditional probability is but I don't know how to find P(first three of these four samples each has exactly two people with blood group A and total number from four samples having blood group A is 9) and P(total number having blood group A is 9). Please help!

Thank you!

2. Originally Posted by Tangera

Q: In a large population, the proportion having blood group A is 35%. Specimens of blood from the first five people attending a clinic are to be tested. It can be assumed that these five people are a random sample from the population. The random variable X denotes the number of people in the sample who are found to have blood group A.

(iii) Find the probability that the first three of these four samples each has exactly two people with blood group A, given that the total number from these four samples having blood group A is 9.

===
Hi! Below is my suggestion: correct me if I'm wrong

Just look at the question in a simpler way: Given that total number of blood A is 9, and exactly two people in the first three of four samples.
That will mean the the fourth sample has 3 blood A people. Hence,

$P(X=2)={5 \choose 2}(\frac{35}{100})^2(\frac{65}{100})^3=0.3364$

$P(X=3)={5 \choose 3}(\frac{35}{100})^3(\frac{65}{100})^2=0.1811$

Answer= $\frac{1}{4}(0.3364)^3(0.1811)$

3. Hello!

4. Originally Posted by noob mathematician
Hi! Below is my suggestion: correct me if I'm wrong

Just look at the question in a simpler way: Given that total number of blood A is 9, and exactly two people in the first three of four samples.
That will mean the the fourth sample has 3 blood A people. Hence,

$P(X=2)={5 \choose 2}(\frac{35}{100})^2(\frac{65}{100})^3=0.3364$

$P(X=3)={5 \choose 3}(\frac{35}{100})^3(\frac{65}{100})^2=0.1811$

Answer= $\frac{1}{4}(0.3364)^3(0.1811)$
Oic.. haha.. so it's a conditional probability.
$P(B)=P(\textrm{9 people out of 20 people with blood A})={20 \choose 9}(\frac{35}{100})^9(\frac{65}{100})^{11}=0.1158$

$P(A\cap B)=(0.3364)^3(0.1811)$

Answer: $P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{(0.3364)^3(0.1811)}{0.1158}=0.0595$

5. Okay now I understand! Thank you for your help! ^^