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  Suppose that a random sample of 10 people are selected for a trial of a vaccine for a childhood disease.

(a) If 40% of the population has had the disease, what is the probability of at least 3 in the sample who have had the disease?

Solution:Let X be the number in the sample who have had the disease. Then $X \sim Bin(10,.4)$.

$$P[X \ge 3] \; = \; 1-P[X \le 2] \; = \; 1- 0.167 \; = \; 0.833 .$$

(b) Suppose 80% of the population will develop a rash from the vaccination. What is the probability that 5 or fewer subjects develop a rash?

Solution:Let X be the number in the sample who develop a rash. Then $X \sim Bin(10,.8)$. We are not given values when p > .5, so turn things around and consider Y, the number in the sample who do not develop a rash. Then $Y \sim Bin(10,.2)$. Also the event $[X \le 5]$ is the same $[Y \ge 5]$ since if x =5 develop a rash, then y=5 don't develop a rash, and if x=4 develop a rash, then y=6 don't develop a rash, etc.

$$P[Y \ge 5] \; = \; 1-P[Y \le 5] \; = \; 1- 0.994 \; = \; 0.006 .$$

The last displayed equation has an error -- it should be 1-P[Y<=4] = 1- 0.967 = 0.033.  Sorry for the mistake.

Use the tabulated values of the Binomial distribution function to answer the questions.
 
 

x p=.1 p=.2 p=.3 p=.4 p=.5

0.349 0.107 0.028 0.006 0.001

1 0.736 0.376 0.149 0.046 0.011

2 0.930 0.678 0.383 0.167 0.055

3 0.987 0.879 0.650 0.382 0.172

4 0.998 0.967 0.850 0.633 0.377

5 1.000 0.994 0.953 0.834 0.623

6 1.000 0.999 0.989 0.945 0.828

7 1.000 1.000 0.998 0.988 0.945

8 1.000 1.000 1.000 0.998 0.989

9 1.000 1.000 1.000 1.000 0.999

10 1.000 1.000 1.000 1.000 1.000

Problems? Questions? (I may have made a mistake, so don't hestitate to ask) dcox@stat.rice.edu
3/12/2001