A medical survey was conducted in order to establish the proportion of the population which was infected along with cancer. The results indicated that 40 percent of the population was suffering from the disease.
A sample of 6 people was later taken and examined for the disease. Determine the probability that the given outcomes were observed
a) Merely one person had the disease
b) Exactly two people had the disease
c) Mostly two people had the disease
d) At least two people had the disease
e) Three or four(4) people had the disease
Solution
P(a persona having cancer) = 40% = 0.4 = P
P(a person not having cancer) = 60% = 0.6 = 1 - p = q
a) P(only one person having cancer)
= 6C1 (0.4)(0.6)5
= 6!/(5! 1!)(0.4)1(0.6)5
= 0.1866
Note that from the formula
nCrprqn-r: where as: n = sample size = 6
p = 0.4
r = 1 = simply one person having cancer
b) P(2 people had the disease)
= 6C2 (0.4)2 (0.6)4
6!/(6! 2!)= (0.4)2 (0.6)5
(6 * 5 * 4!)/(4! * 2 *1)= (0.4)2 (0.6)5
= 15 × (0.4)2 (0.6)5
= 0.311
c) P(at most 2) = P(0) + P(1) + P(2) = P(0) or P(1) or P(2)
So we estimate the probability of each and add them up.
P(0) = P(nobody having cancer)
= 6C0 (0.4)0(0.6)6
6!/(0! 6!)= (0.4)0(0.6)6
= (0.6)6
= 0.0467
The probabilities of P(1) and P(2) have been worked out in part (a) and
Hence P(at most 2) = 0.0467 + 0.1866 + 0.311 = 0.5443
d) P(at least 2)
= P(2) + P(3) + P(4) + P(5) + P(6)
= 1 - [P(0) + P(1)] it is a shorter way of working out the solution as
[P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1]
= 1 - (0.0467 + 0.1866)
= 0.7667
e) P(3 or 4 people had the disease)
= P(3) +P(4)
= 6C3(0.4)3(0.6)3 + 6C4(0.4)4(0.6)2
= ( 6!/(3! 3!)) (0.4)3(0.6)3 + (6!/(2! 4!)) (0.4)4(0.6)2
= {(6 × 5 × 4 × 3!)/ (3 × 2 × 1 × 3!)}(0.4)3(0.6)3 + {(6 × 5 × 4!)/(2 × 1 × 4!)} (0.4)4(0.6)2
= 20(0.4)3(0.6)3 + 15(0.4)4(0.6)2
= (20 × 0.013824) + (15 × 0.009216)
= 0.27648 + 0.13824
= 0.41472