Determine the probability , Mathematics

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

Posted Date: 2/19/2013 8:15:35 AM | Location : United States







Related Discussions:- Determine the probability , Assignment Help, Ask Question on Determine the probability , Get Answer, Expert's Help, Determine the probability Discussions

Write discussion on Determine the probability
Your posts are moderated
Related Questions
Patio measures 24 meters square. Patio stone are 30 cm each side. How many stones are required to cover the patio?

Classify the following discrete-time signals as energy or power signals. If the signal is of energy type, find its energy. Otherwise, find the average power of the signal. X 1

A 125-foot tower is located on the side of a mountain that is inclined at 32° to the horizontal. A guy wire is to be fitted to the top of the tower and anchored at a point 55 feet


How t determine locus of a goven point

The alternative hypothesis When formulating a null hypothesis we also consider the fact that the belief may be found to be untrue thus we will refuse it.  Therefore we formula

The base and corresponding altitude of a parallelogram are 10 cm and 12 cm reap. If the other altitude is 8 cm , find the length of the other pair of parallel side

what''s the main purpose of algebra in our daily life

If 7 cosec?-3cot? = 7, prove that 7cot? - 3cosec? = 3. Ans:    7 Cosec?-2Cot?=7 P.T 7Cot? - 3 Cosec?=3 7 Cosec?-3Cot?=7 ⇒7Cosec?-7=3Cot? ⇒7(Cosec?-1)=3Cot? ⇒7(C

Consider this. You have four units A, B, C and D. You are asked to select two out of these four units. How do you go about this particular task? Will your methodo