Binomial distribution

**Tangera** · May 4th 2009, 05:14 AM

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.

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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!

**noob mathematician** · May 5th 2009, 07:00 AM

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

**Tangera** · May 5th 2009, 11:12 PM

Hello!

Thank you for your suggestion! I had tried it out your method beforehand but I could not get the answer given to me, which is 0.0595. I am still not sure how to get it...Please help!

**noob mathematician** · May 6th 2009, 12:18 AM

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

**Tangera** · May 7th 2009, 12:29 AM

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

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