Last week a gumball machine dispensed 10 white gumballs and 16 other gumballs. Considering this data, how many of the next 78 gumballs dispensed should you expect to be white gumballs?

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romancemrf · Accepted Answer

Step 1  Introduction:  If X∼B∈omial(n,p), then the probability mass function of X is:  p(x)={(nx)pxqn−x;x=0,1,...,n;0&amp;lt;p&amp;lt;1;q=1−p0;otherwise.  Here, n is the number of independent trials, p is the probability of success in each trial, and q is the probability of failure.  The expectation of X is, E(X)=np.  Step 2  Calculation:  Denote X as the number of white gumballs to be dispensed by the machine, out of the 78 total gumballs that it will dispense.  It may be assumed that the colors of the gumballs dispensed are independent of one another. Thus, there may be considered to be n=78 independent trials, or gumballs dispensed.  It is given that in the previous week, 10 white gumballs, and 16 gumballs of some other color were dispensed. This implies that 10 of the 26(=10+16) dispensed gumballs were white. If obtaining a white gumball is considered as a success and the number of successes obtained in the previous week is representative of the number of successes in general, then the probability of success is, p=1026=513, which is constant for all the gumballs dispensed.  Therefore, it may be assumed that X has a binomial distribution with parameters, n=78,p=513.  As a result, the expected number of successes, that is, the expected number of white gumballs out of the 78 gumballs dispensed will be:  E(X)  =np  =78×(513)  =30.  Thus, 30 of the next 78 gumballs dispensed may be expected to be white.

Last week a gumball machine dispensed 10 white gumballs and

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