Question

Computer keyboard failures can be attributed to electrical defects or mechanical defects. A repair facility currently has 25 failed keyboards, 7 of which have electrical defects and 18 of which have mechanical defects.(a) How many ways are there to randomly select 6 of these keyboards for a thorough inspection (without regard to order)?(b) In how many ways can a sample of 6 keyboards be selected so that exactly three have a mechanical defect?(c) If a sample of 6 keyboards is randomly selected, what is the probability that no more than 2 of these will have an electrical defect?

Answer 1

Answer:

The number of ways to select 6 keyboards is 177100.

The number of ways so that exactly three have a mechanical defect is 28560.

The probability that no more than 2 electrical defect is 0.806.

Step-by-step explanation:

Consider the provided information.

A repair facility currently has 25 failed keyboards, 7 of which have electrical defects and 18 of which have mechanical defects.

Part(A)

It is given that we need to find How many ways are there to randomly select 6 of these keyboards for a thorough inspection.

We need to select 6 keyboards out of 25.

Thus, the number of ways are:

$_{25}C_6=\frac{25!}{(25-6)!6!}$

$_{25}C_6=\frac{25!}{19!6!}$

$_{25}C_6=177100$

Hence, the number of ways to select 6 keyboards is 177100.

Part(B)

In how many ways can a sample of 6 keyboards be selected so that exactly three have a mechanical defect?

We need to select 6 keyboards if exactly three have a mechanical defect that means the remain 3 have a electrical defect.

3 keyboard with electrical defects: $_{7}C_3$

3 keyboard with mechanical defects: $_{18}C_3$

This can be written as:

$_{7}C_3\times_{18}C_3=35\times816$

$_{7}C_3\times _{18}C_3=28560$

Thus, the number of ways so that exactly three have a mechanical defect is 28560.

Part(C)

If a sample of 6 keyboards is randomly selected, what is the probability that no more than 2 of these will have an electrical defect?

No more than 2 of these will have an electrical defect that means the number of electrical defect keyboards can be 0, 1 or 2.

P(No more than 2 electrical defect)={P(0 electrical defect)+P(1 electrical defect)+P(2 electrical defect)}

This can be written as:

$\frac{_{7}C_0\times _{18}C_6}{_{25}C_6}+\frac{_{7}C_1\times _{18}C_5}{_{25}C_6}+\frac{_{7}C_2\times _{18}C_4}{_{25}C_6}$

$\frac{\left(18564+7\cdot8568+21\cdot3060\right)}{177100}$

$0.806324110672\approx0.806$

Hence, the probability that no more than 2 electrical defect is 0.806.