Answer to Question #158322 in Statistics and Probability for Dp

Question

A lot consists of 10 good articles, 4 articles with minor defects and 2 with major defects. Two articles are chosen at random from the lot (without replacement) Find the probability that: (a) both are good(b) both has major defects, (c) at least 1 is good (d) neither has major defects, (e) neither is good

Answer 1

There are total 16 articles out of which 10 good articles, 4 articles with minor defects and 2 articles with major defects.

Two articles can be chosen in ${^{16}C_2}$ ways. Therefore the sample space contains $^{16}C_2$ elements.

(a) Let $A$ be an event such that both the articles are good.

Then the number of elements in favour of $A$ is ${^{10}C_2}$.

$\therefore$ $P(A)=\frac{^{10}C_2}{^{16}C_2}=\frac{45}{120}=\frac{3}{8}$

(b) Let $B$ be an event such that both the articles have major defects

Then the number of elements in favour of $B$ is ${^{2}C_2}$

$\therefore$ $P(B)=\frac{^{2}C_2}{^{16}C_2}=\frac{1}{120}$

(c) Let $C$ be an event such that at least 1 article is good. Then two cases arise.

Case 1: It will be 1 good article and 1 article with major or minor defects

Case 2: 2 articles are good.

For the first case, 1 good article and 1 article with major or minor defects can be chosen in ${^{10}C_1}×{^{6}C_1}$ ways .

For second case , 2 good articles can be chosen in ${^{10}C_2}$ ways.

So total number of elements in favour of $C$ is $({^{10}C_1}×{^{6}C_1})+{^{10}C_2}=(60+45)=105$

$\therefore$ $P(C)=\frac {105}{120}=\frac{7}{8}$

(d) Let $D$ be an event such that neither has major defects. Then articles will be either good or have minor defects.

Then the number of elements in favour of $D$ is ${^{14}C_2}$

$\therefore$ $P(D)=\frac {^{14}C_2}{^{16}C_2}=\frac{91}{120}$

(e) Let $E$ be an event such that neither is a good article. Then articles will be either major or minor defects.

Then the number of elements in favour of $E$ is ${^6C_2}$

$\therefore$ $P(E)=\frac{^6C_2}{^{16}C_2}=\frac{15}{120}=\frac{1}{8}$