Question

A box of 60 remote control devices contains 4 that have a defective power button. If devices are randomly sampled from the box and inspected one at a​ time, determine the following probabilities. a. The probability that the first control device is defective. b. The probability that the first control device is good and the second control device is defective. c. The probability that the first three sampled devices are all good. a. The probability that the first control device is defective is nothing. ​(Round to four decimal places as​ needed.)

Answer 1

Answer with Step-by-step explanation:

The probability that the device is good drawn in first try

`P(E)=(4)/(60)=(1)/(15)`

Probability that the device drawn is defective in first try is

`P(E_2)=(56)/(60)=(14)/(15)`

Part a)

Probability that the device obtained in first try is defective is 1/15.

Part b)

Probability that the device obtained in first try is good and the second is defective is

`P(E')=P(E_2)* P(E_3)`

where

`P(E_3)` is the probability that the second is defective provided first is good is
`(4)/(59)`

Thus

`P(E')=(56)/(60)* (4)/(59)=0.0632`

Part c)

Probability that all the three samples are good are

Probability that first device is good times

Probability that second device is also good times

Probability that third device is good

`P(E)=(56)/(60)* (55)/(59)* (54)/(58)=0.81`