Question

A bag contains 7 green marbles, 2 white marbles, 5 orange marbles, 9 green marbles, and 7 red marbles. A marble will be drawn from the bag and replaced 180 times. What is a reasonable prediction for the number of times a white or red marble will be drawn?.

Answer 1

Prediction for some count is usually expressible as expectation of some random variable. The reasonable prediction for the number of times a white or red marble will be drawn is 54

What does the expectation of a random variable tells about?

A  random variable is usually taking some numerical information of each elementary event of the sample space . (map from sample space to real numbers). Its expected value is the best prediction value in most of the cases.

How to find that a given condition can be modeled by binomial distribution?

Binomial distributions consists of n independent Bernoulli trials.

Bernoulli trials are those trials which end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

Suppose we have random variable X pertaining binomial distribution with parameters n and p, then it is written as

$X \sim B(n,p)$

The probability that out of n trials, there'd be x successes is given by

$P(X =x) = \: ^nC_xp^x(1-p)^{n-x}$

For the given case, as the ball being drawn is being replaced too, so we can take all 180 experiments to be independent from each other.

If we call

Success = Event of drawing white or  red marble

Failure = Event of drawing any other marble instead of white or red

Then, each marble drawing event is having two outcomes: success or failure. This makes each drawing a Bernoulli  experiment.

Modelling this situation with binomial distribution would be good.

Let X = number of successes in those 180 experiments.

• P(Success) = 9/30 = 0.3 = p(say) (total 9 marbles in favor, so 9 ways of drawing single marble, and total 30 marbles are there, and thus 30 ways of drawing single marble out of them)
• P(Failure) = P(Not success) = 1-p = 0.7 = q (say)

Thus, we get

$X \sim B(n = 180, p= 0.3)$

The expectation of random variable X is the best prediction of count of times we will get in those 180 trials.

The expectation of a  random variable is: $E(X) = np$

Thus, we get:

$E(X) = 180 \times 0.3 = 54$

Thus, the reasonable prediction for the number of times a white or red marble will be drawn is 54

Learn more about binomial distribution here:

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