Nancy dancing lesson is at 9:00AM
with 0.06 probability that she catch it.
The expected number of lessons she might catch is 15.
Find the probability that she catches the first 5 lessons.
Hmmmm. Prob of .06 that she gets the first lesson, but what is the probably for later lessons? Are the lessons sequential? Like one per hour?
You have a time issue here, honestly I do not see how this is a binomial distribution.
catch means attend
To find the probability that Nancy catches the first 5 lessons, we can use the concept of binomial distribution.
The binomial distribution is used to calculate the probability of getting a certain number of successes in a fixed number of independent Bernoulli trials (where each trial has only two possible outcomes: success or failure).
In this case, the probability of success is 0.06 (Nancy catching a lesson) and the number of trials is 15 (the expected number of lessons).
To calculate the probability of catching the first 5 lessons, we need to calculate the probability of exactly 5 successes in 15 trials:
P(X = 5) = (15 choose 5) * (0.06)^5 * (1-0.06)^(15-5)
In this formula, "n choose k" represents the number of ways to choose k items from a set of n items, and is calculated as:
(15 choose 5) = 15! / (5! * (15-5)!)
Substituting the values into the formula, we get:
P(X = 5) = (15! / (5! * 10!)) * (0.06)^5 * (0.94)^10
Calculating the combination and simplifying the equation, we have:
P(X = 5) = (15 * 14 * 13 * 12 * 11 / 5 * 4 * 3 * 2 * 1) * (0.06)^5 * (0.94)^10
P(X = 5) = (3003) * (0.006084) * (0.1756)
P(X = 5) = 0.32070528
Therefore, the probability that Nancy catches the first 5 lessons is approximately 0.3207 or 32.07%.