Question

three computers are chosen at random from an inventory of dell and acer computers for a bookstore display. assume the same number each brand of computers is in stock. find the probability that
(a) all three will be dells.
(b) exactly two will be acers.
(c) at most two will be acers.
write your answers in exact, simplified form.
(a) the probability that all three will be dells is \\( \frac { 1 } { 8 } \\).
(b) the probability that exactly two will be acers is \\( \square \\).

Explanation:

Step1: Calculate the number of ways to choose 3 computers

Assume there are \(n\) Dell and \(n\) Acer computers. The total number of computers is \(2n\). The number of ways to choose 3 computers out of \(2n\) is given by the combination formula \(C(2n,3)=\frac{(2n)!}{3!(2n - 3)!}=\frac{2n(2n-1)(2n - 2)}{6}\)

Since the probability that all three are Dells is \(\frac{1}{8}\), and the number of ways to choose 3 Dells out of \(n\) is \(C(n,3)=\frac{n!}{3!(n - 3)!}=\frac{n(n - 1)(n - 2)}{6}\)

We have \(\frac{C(n,3)}{C(2n,3)}=\frac{1}{8}\)

Substitute the combination formulas:

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This is wrong. Let's assume there are \(4\) Dell and \(4\) Acer computers (\(n = 4\), total \(N=8\))

Step2: Calculate part (b)

The number of ways to choose exactly 2 Acers (and 1 Dell) is \(C(4,2)\times C(4,1)\)

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The number of ways to choose 3 computers out of \(8\) is \(C(8,3)=\frac{8!}{3!(8 - 3)!}=\frac{8\times7\times6}{3\times2\times1}=56\)

The probability \(P=\frac{24}{56}=\frac{3}{7}\)

Answer:

\(\frac{3}{7}\)