Question

We have a binomial experiment with n = 18 trials, each with probability p = 0.15 of a success.

A success occurs if a student withdraws from a class, so the number of successes, x, will take on the values 0, 1, and 2. The probability of each x value, denoted f(x), can be found using a table like the one below. Note that these values are rounded to four decimal places.

n	x	p
		0.10	0.15	0.20	0.25
18	0	0.1501	0.0536	0.0180	0.0056
	1	0.3002	0.1704	0.0811	0.0338
	2	0.2835	0.2556	0.1723	0.0958
	3	0.1680	0.2406	0.2297	0.1704
	4	0.0700	0.1592	0.2153	0.2130
	5	0.0218	0.0787	0.1507	0.1988

For an experiment with n = 18 trials, the probability of exactly x = 0 successes where the probability of a success on a trial is p = 0.15 can be found by going along the row for x = 0 within the n = 18 grouping until you get to the column for p = 0.15. Doing so gives f(0) =  .

Use the above table to find the probabilities for x = 1 success, f(1), and x = 2 successes, f(2).

f(1)	=
f(2)	=

Answer 1

From the given information X follows binomial distribution with following parameyters.

n = 18, p = 0.15

Also we have given the table for n = 18 and different values of p as 0.10, 0.15, 0.20 and 0.25

This is the table of exact binomial probabilities.

First we want to find f(0) = P(X = 0) for n = 18, and p = 0.15

It is correspond to the row of heading " 0 " and column of heading " 0.15 "

Therefore, f(0) = 0.0536

Look the following image:

Now let's find f(1)

f(1) = P(X = 1) for n = 18, and p = 0.15

It is correspond to the row of heading " 1 " and column of heading " 0.15 "

Therefore, f(1) = 0.1704

Look the following image:

Now let's find f(2)

f(2) = P(X = 2) for n = 18, and p = 0.15

It is correspond to the row of heading " 2 " and column of heading " 0.15 "

Therefore, f(2) = 0.2556

Look the following image:

Therefore,

f(0) = 0.0536

f(1) = 0.1704

f(2) = 0.2556