A company markets educational software products, and is ready to place three new products on the market. Past experience has shown that for this particular software, the chance of "success" is 80%. Assume that the probability of success is independent for each product. What is the probability that exactly 1 of the 3 products is successful?

Use a binomial probability formula or use a binomial probability table.

Formula:

P(x) = (nCx)(p^x)[q^(n-x)]

P(1) = (3C1)(.8^1)(.2^2)

I'll let you finish the calculation.

If you use the table, p = .8 (for 80%), n = 3 (sample size), x = 1.

(Note: q in the formula is 1-p or .2).

I hope this will help.

.096

To find the probability that exactly 1 of the 3 products is successful, we can use the binomial probability formula. The formula is:

P(X=k) = nCk * p^k * (1-p)^(n-k)

Where:
- P(X=k) is the probability of getting exactly k successes.
- n is the total number of trials or products (in this case, 3).
- k is the number of successes we want (in this case, 1).
- p is the probability of success for each trial (in this case, 0.80).

So, let's plug in the values into the formula:

P(X=1) = 3C1 * 0.80^1 * (1-0.80)^(3-1)

Now, let's calculate each term:

3C1 = 3! / (1! * (3-1)!) = 3
0.80^1 = 0.80
(1-0.80)^(3-1) = 0.20^2 = 0.04

Now, substitute the values back into the formula:

P(X=1) = 3 * 0.80 * 0.04 = 0.096

So, the probability that exactly 1 of the 3 products is successful is 0.096, or 9.6%.

6.36