Question

A binomial probability experiment is conducted with the given parameters. Compute the probability of x successes in the n independent trials of the experiment.P=0.6. n = 9 x <= 3

Answer 1

The student's question involves calculating the probability of 3 or fewer successes in a binomial probability experiment with 9 trials and a success probability of 0.6. The sum of probabilities for exactly 0, 1, 2, and 3 successes will provide the answer.

Step-by-step explanation:

Calculating Binomial Probabilities-

The problem presents a binomial probability experiment where the number of trials, n, is 9, and the probability of success, p, is 0.6. The question asks us to find the probability of obtaining x successes where x is less than or equal to 3. To solve this, we would use the binomial probability formula:

P(X = x) = (nCx) * px * (1-p)n-x

Where nCx is the combination of n items taken x at a time.

Since we are looking for the probability of getting 3 or fewer successes (x ≤ 3), we need to add the probabilities of getting exactly 0, 1, 2, and 3 successes. Computing each of these probabilities separately and summing them up will give us the required probability for the given conditions of the experiment.

Answer 2

To compute the probability of x successes in n independent trials of a binomial probability experiment, we use the binomial probability formula:

P(x) = (n C x) * p^x * (1-p)^(n-x)

In this case, we have the following parameters:

- Probability of success, p = 0.6

- Number of independent trials, n = 9

- We want to compute the probability of x successes, where x <= 3

Now, let's calculate the probabilities for each value of x <= 3:

For x = 0:

P(0) = (9 C 0) * 0.6^0 * (1-0.6)^(9-0)

= (1) * 1 * 0.4^9

= 0.4^9

≈ 0.001572864

For x = 1:

P(1) = (9 C 1) * 0.6^1 * (1-0.6)^(9-1)

= (9) * 0.6 * 0.4^8

≈ 0.04023328

For x = 2:

P(2) = (9 C 2) * 0.6^2 * (1-0.6)^(9-2)

= (36) * 0.6^2 * 0.4^7

≈ 0.12069984

For x = 3:

P(3) = (9 C 3) * 0.6^3 * (1-0.6)^(9-3)

= (84) * 0.6^3 * 0.4^6

≈ 0.214990848

To compute the probability of x <= 3, we sum up the probabilities for each value of x:

P(x <= 3) = P(0) + P(1) + P(2) + P(3)

≈ 0.001572864 + 0.04023328 + 0.12069984 + 0.214990848

≈ 0.377496832

Therefore, the probability of having x <= 3 successes in 9 independent trials, where the probability of success is 0.6, is approximately 0.3775.