Answer:
a. 1365 ways
b. Probability = 0.4096
c. Probability = 0.5904
Explanation:
Given
PCs = 15
Purchase = 3
Solving (a): Ways to select 4 computers out of 15, we make use of Combination formula as follows;
[tex]^nC_r = \frac{n!}{(n-r)!r!}[/tex]
Where [tex]n = 15\ and\ r = 4[/tex]
[tex]^{15}C_4 = \frac{15!}{(15-4)!4!}[/tex]
[tex]^{15}C_4 = \frac{15!}{11!4!}[/tex]
[tex]^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}[/tex]
[tex]^{15}C_4 = \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1}[/tex]
[tex]^{15}C_4 = \frac{32760}{24}[/tex]
[tex]^{15}C_4 = 1365[/tex]
Hence, there are 1365 ways
Solving (b): The probability that exactly 1 will be defective (from the selected 4)
First, we calculate the probability of a PC being defective (p) and probability of a PC not being defective (q)
From the given parameters; 3 out of 15 is detective;
So;
[tex]p = 3/15[/tex]
[tex]p = 0.2[/tex]
[tex]q = 1 - p[/tex]
[tex]q = 1 - 0.2[/tex]
[tex]q = 0.8[/tex]
Solving further using binomial;
[tex](p + q)^n = p^n + ^nC_1p^{n-1}q + ^nC_2p^{n-2}q^2 + .....+q^n[/tex]
Where n = 4
For the probability that exactly 1 out of 4 will be defective, we make use of
[tex]Probability = ^nC_3pq^3[/tex]
Substitute 4 for n, 0.2 for p and 0.8 for q
[tex]Probability = ^4C_3 * 0.2 * 0.8^3[/tex]
[tex]Probability = \frac{4!}{3!1!} * 0.2 * 0.8^3[/tex]
[tex]Probability = 4 * 0.2 * 0.8^3[/tex]
[tex]Probability = 0.4096[/tex]
Solving (c): Probability that at least one is defective;
In probability, opposite probability sums to 1;
Hence;
Probability that at least one is defective + Probability that at none is defective = 1
Probability that none is defective is calculated as thus;
[tex]Probability = q^n[/tex]
Substitute 4 for n and 0.8 for q
[tex]Probability = 0.8^4[/tex]
[tex]Probability = 0.4096[/tex]
Substitute 0.4096 for Probability that at none is defective
Probability that at least one is defective + 0.4096= 1
Collect Like Terms
Probability = 1 - 0.4096
Probability = 0.5904
