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

Question 1120948: A medical survey was conducted in order to establish the proportion which was infected with cancer. The results indicated that 40% of the population were suffering from cancer.A sample of 6 people was later taken and examined for the disease. Find the probability that the following outcomes were observed:
(i) Only one person had cancer
(ii) At most two people had cancer
(iii) At least two people had cancer
(iv) Three or four people had cancer

Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
this problem can be viewed as a binomial experiment, since there are two possible outcomes for the patient and the probability (0.40) stays constant
:
binomial probability formula is
:
Probability (P) (k successes in n trials) = nCk * p^k * (1-p)^(n-k), where nCk = n! / (k! * (n-k)!)
:
p = 0.40
:
(i) P (1 success in 6 trials) = 6C1 * (0.40)^1 * (1-0.40)^(6-1) = 0.1866
:
(ii) P (k < or = 2 in 6 trials) = 0.5443
:
(iii) P (k > or = 2 in 6 trials) = 0.7667
:
(iv) P (k = 3 or 4 in 6 trials) = P(k = 3 in 6 trials) + P(k = 4 in 6 trials) = 0.2764 + 0.1382 = 0.4146
:

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