SOLUTION: ) A medical survey was conducted in order to establish the proportion of the population which was infected with cancer. The results indicated that 40% of the population was sufferi

Question 980063: ) A medical survey was conducted in order to establish the proportion of the population which was infected with cancer. The results indicated that 40% of the population was suffering from the disease. A sample of 6 people was later taken and examined for the disease. Find the probability that the following outcomes were observed
a) Only one person had the disease (4 Marks)
b) Exactly two people had the disease (3 Marks)
c) At most two people had the disease (3 Marks)
d) Three or four people had the disease (2 Marks)

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
I'll do the first two parts (a and b) to get you started.

For parts a) through d), the following is true

n = 6 is the sample size

p = 0.40 is the probability of having the disease

The binomial probability distribution fromula will be used. The formula is P(x = k) = (n C k)*(p)^(k)*(1-p)^(n-k). I'm using the combination formula which is (n C r) = (n!)/(r!*(n-r)!) for the "n C k" portion.

In part a), the value of k is k = 1. In part b), the value of k is k = 2.
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a)
Exactly One

k = 1

Binomial distribution

P(x = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(x = 1) = (6 C 1)*(0.40)^(1)*(1-0.40)^(6-1)
P(x = 1) = (6)*(0.40)^(1)*(0.6)^(5)
P(x = 1) = 0.186624

Answer: 0.186624

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b)
Exactly Two

k = 2

Binomial distribution

P(x = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(x = 2) = (6 C 2)*(0.40)^(2)*(1-0.40)^(6-2)
P(x = 2) = (15)*(0.40)^(2)*(0.6)^(4)
P(x = 2) = 0.31104

Answer: 0.31104

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