Question

Use the scenario below to determine the correct values of n, p, q and x of the binomial distribution. Suppose that in a certain video game there is a 2.3% item drop rate of glowing mana after defeating a Baal. Assume 19 Baals are defeated. What is the probability that 2 glowing manas will drop?

Answer 1

The vale of n, p, q and x are 19, 0.023, 0.977, and 2, respectively

The probability that exactly 2 glowing manas will drop after defeating 19 Baals in the video game is approximately 0.061 or 6.1%.

What is the probability that 2 glowing manas will drop?

Given:

Number of trials (defeating Baals), n = 19.

Probability of success (getting a glowing mana drop) in a single trial,

p = 0.023 (2.3% expressed as a decimal).

Probability of failure (not getting a glowing mana drop) in a single trial,

q = 1 - p = 1 - 0.023 = 0.977.

Desired number of successes, x = 2.

The formula for the probability of exactly x successes in n trials using the binomial probability formula is:

P(X = x) = (n choose x) *
`p^x * q^(n - x)`

Substituting the values:

P(X = 2) = (19 choose 2) *
`(0.023)^2 * (0.977)^(19 - 2)`

Calculating:

P(X = 2) = (19! / (2! * (19 - 2)!)) * 0.000529 * 0.673

P(X = 2) = 171 * 0.000529 * 0.673

P(X = 2) ≈ 0.061

Therefore, the probability that exactly 2 glowing manas will drop after defeating 19 Baals in the video game is approximately 0.061 or 6.1%.