Answer to Question #95831 in Statistics and Probability for Jess

Question

Q1
A soft-drink machine dispenses only regular Coke and Diet Coke. Sixty percent of all purchases from this machine are diet drinks. The machine currently has ten cans of each type. If 15 customers want to purchase drinks prior to the machine being restocked, what is the probability that each of the 15 is able to purchase the type of drink desired?

Answer 1

If there are 15 customers, and X represents the number who choose diet, and Y represents the number who choose non-diet, then  X+Y=15. So we only need one variable to describe the number of customers who choose a drink of a particular type.

Since there are only 10 cans of each type available and X+Y = 15, the allowable combinations of (X,Y) are

$(5, 10),(6,9),(7,8),(8,7),(9,6),(10,5)$

The random variable X follows binomial distribution with $p=0.6$ and $n=15$: $X\sim B(15,0.6)$

$b(x;n,p)=\binom{n}{x}p^xq^{n-x}, x=0, 1,2,...,n; \ q=1-p$

The probability that each of to 15 customers get the drink they want is

$P(5\leq X\leq10)=P(X=5)+P(X=6)$ $+P(X=7)+P(X=8)+P(X=9)+P(X=10)$

$P(X=5)=\binom{15}{5}0.6^5(1-0.6)^{15-5}\approx0.02448564$ $P(X=6)=\binom{15}{6}0.6^6(1-0.6)^{15-6}\approx0.06121411$ $P(X=7)=\binom{15}{7}0.6^7(1-0.6)^{15-7}\approx0.11805577$ $P(X=8)=\binom{15}{8}0.6^8(1-0.6)^{15-8}\approx0.17708366$ $P(X=9)=\binom{15}{9}0.6^9(1-0.6)^{15-9}\approx 0.20659761$ $P(X=10)=\binom{15}{10}0.6^{10}(1-0.6)^{15-10}\approx 0.18593784$

$P(5\leq X\leq10)\approx0.02448564+0.06121411+0.11805577+$ $+0.17708366+0.20659761+0.18593784\approx$ $\approx0.773375\ (to\ 6\ decimal\ places)$

The probability that each of the 15 is able to purchase the type of drink desired is $0.773375$ (about $77.3\%$).