Question

Compute the probability of X successes, using the binomial distribution table. Part 1 of 4 (a) n=5,p=0.3,X=3 P(X)= Part 2 of 4 (b) n=7,p=0.7,X=6 P(X)= Part 3 of 4 (c) n=15,p=0.2,X=4 P(X)= Part 4 of 4 (d) n=18,p=0.6,X=7

Answer 1

The probabilities for the binomial distribution problems can be found using the binompdf function or a binomial distribution table, focusing on the given number of trials, probability of success, and exact count of successes.

Step-by-step explanation:

Understanding Binomial Distribution for Calculations

To compute the probabilities for different values of X using the binomial distribution, one could either refer to a standard binomial distribution table or use technology like a graphing calculator. For the specific problems given:

• For (a) n=5, p=0.3, X=3, we use the function binompdf(5, 0.3, 3) to find the probability.
• For (b) n=7, p=0.7, X=6, we use the function binompdf(7, 0.7, 6) to find the probability.
• For (c) n=15, p=0.2, X=4, we use the function binompdf(15, 0.2, 4) to find the probability.
• For (d) n=18, p=0.6, X=7, we use the function binompdf(18, 0.6, 7) to find the probability.

If a calculator or software is not accessible, one would use a binomial distribution table matching the values of n, p, and X to find the probability.

Remember that X must be a whole number, and the probability is found for that exact count of successes in n trials.

Answer 2

The probabilities are 0.1323, 0.247, 0.185 and 0.98.

The probability of obtaining a specific number of successes (X) in a binomial distribution can be calculated using the formula:

`\[ P(X) = \binom{n}{X} * p^X * (1 - p)^(n - X) \]`

Let's calculate each probability:

(a) For n = 5, p = 0.3, and X = 3:

`\[ P(X) = \binom{5}{3} * 0.3^3 * (1 - 0.3)^(5 - 3) \]`

P(X) = 10 x 0.027 x 0.49

P(X) = 0.1323

(b) For n = 7, p = 0.7, and X = 6:

`\[ P(X) = \binom{7}{6} * 0.7^6 * (1 - 0.7)^(7 - 6) \]`

P(X) = 7 x 0.117649 x 0.3

P(X) = 0.247

(c) For n = 15, p = 0.2, and X = 4:

`\[ P(X) = \binom{15}{4} * 0.2^4 * (1 - 0.2)^(15 - 4) \]`

P(X) = 1365 x 0.0016 x 0.1073741824

P(X) = 0.185

(d) For n = 18, p = 0.6, and X = 7:

`\[ P(X) = \binom{18}{7} * 0.6^7 * (1 - 0.6)^(18 - 7) \]`

P(X) = 31824 x 0.027 x 0.1073741824

P(X) = 0.98

These are the calculated probabilities for each scenario (a), (b), (c), and (d) using the binomial distribution formula.