Question

A boxcar contains six complex electronic systems. Two of the six are to be randomly selected for thorough testing and then classified as defective or not defective.
a. If two of the six systems are actually defective, find the probability that at least one of the two systems tested will be defective. Find the probability that both are defective.
b. If four of the six systems are actually defective, find the probabilities indicated in part (a).

Answer 1

Answer:

(a)P(At least one defective)$=0.6$

P(Both are defective)$=0.067$

(b)P(At least one defective)$=14/15$

P(Both are defective)$=0.4$

Step-by-step explanation:

We are given that

Total number of complex electronic system, n=6

(a)Defective items=2

Non-defective items=6-2=4

We have to find the  probability that at least one of the two systems tested will be defective.

P(At least one defective)=$\frac{2C_1\times 4C_1}{6C_2}+\frac{2C_2\times 4C_0}{6C_2}$

Using the formula

$P(E)=\frac{favorable\;cases}{total\;number\;of\;cases}$

P(At least one defective)$=\frac{\frac{2!}{1!1!}\times \frac{4!}{1!3!} }{\frac{6!}{2!4!}}+\frac{\frac{2!}{0!2!}\times \frac{4!}{4!}}{\frac{6!}{2!4!}}$

Using the formula

$nC_r=\frac{n!}{r!(n-r)!}$

P(At least one defective)$=\frac{2\times \frac{4\times 3!}{3!}}{\frac{6\times 5\times 4!}{2\times 1\times 4!}}+\frac{1}{\frac{6\times 5\times 4!}{2\times 1\times 4!}}$

P(At least one defective)$=\frac{2\times 4}{3\times 5}+\frac{1}{3\times 5}$

P(At least one defective)$=\frac{8}{15}+\frac{1}{15}=\frac{9}{15}$

P(At least one defective)$=\frac{3}{5}=0.6$

Now, the probability that both are defective

P(Both are defective)=$\frac{2C_2\times 4C_0}{6C_2}$

P(Both are defective)=$\frac{\frac{2!}{0!2!}\times \frac{4!}{4!}}{\frac{6!}{2!4!}}$

P(Both are defective)$=\frac{1}{\frac{6\times 5\times 4!}{2\times 1\times 4!}}$

P(Both are defective)$=\frac{1}{3\times 5}$

P(Both are defective)$=0.067$

(b)

Defective items=4

Non- defective item=6-4=2

P(At least one defective)=$\frac{4C_1\times 2C_1}{6C_2}+\frac{4C_2\times 2C_0}{6C_2}$

P(At least one defective)$=\frac{\frac{4!}{1!3!}\times \frac{2!}{1!1!} }{\frac{6!}{2!4!}}+\frac{\frac{4!}{2!2!}\times \frac{2!}{2!}}{\frac{6!}{2!4!}}$

P(At least one defective)$=\frac{2\times \frac{4\times 3!}{3!}}{\frac{6\times 5\times 4!}{2\times 1\times 4!}}+\frac{\frac{4\times 3\times 2!}{2!\times 2\times 1}}{\frac{6\times 5\times 4!}{2\times 1\times 4!}}$

P(At least one defective)$=\frac{2\times 4}{3\times 5}+\frac{2\times 3}{3\times 5}$

P(At least one defective)$=\frac{8}{15}+\frac{6}{15}=\frac{8+6}{15}$

P(At least one defective)$=\frac{14}{15}$

P(Both are defective)$=\frac{4C_2\times 2C_0}{6C_2}$

P(Both are defective)$=\frac{\frac{4\times 3\times 2!}{2\times 1\times 2!}\times \frac{2!}{2!}}{\frac{6\times 5\times 4!}{2\times 1\times 4!}}$

P(Both are defective)$=\frac{\frac{4\times 3\times 2\times 1}{2\times 1\times 2\times 1}}{3\times 5}$

P(Both are defective)$=\frac{6}{15}=0.4$

P(Both are defective)$=0.4$

Answer 2

Answer:

Step-by-step explanation:

Number of electronic systems = 6

(a) Number of defected systems = 2

Probability of getting at least one system is defective

1 defective and 1 non defective + 2 defective

= (2 C 1 ) x (4 C 1) + (2 C 2) / (6 C 2)

= 3 / 5

(b) four defective

Probability of getting at least one system is defective

2 defective and 2 non defective + 3 defective and 1 non defective + 4 defective  

= (4 C 2 ) x (2 C 2) + (4 C 3 )(2 C 1) + (4 C 4) / (6 C 4)

= 1