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Answer:
a) Let X the random variable of interest, on this case we know that:
This random variable represent that two people have the same birthday in just one classroom
b) We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:
And we can find the individual probability:
And then:
And since we want the probability in the 3 classes we can assume independence and we got:
So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992
Step-by-step explanation:
Previous concepts
A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
The probability mass function for the Binomial distribution is given as:
Where (nCx) means combinatory and it's given by this formula:
Solution to the problem
Part a
Let X the random variable of interest, on this case we know that:
This random variable represent that two people have the same birthday in just one classroom
Part b
We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:
And we can find the individual probability:
And then:
And since we want the probability in the 3 classes we can assume independence and we got:
So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992
and is all you can simplify from it.
so... all we did, was rationaliize it, namely, "getting rid of the pesky radical at the bottom", we do so by simply multiplying it by something that will raise the radicand, to the same degree as the root, thus the radicand comes out.