Question

In: Statistics and Probability

A lot of 106 semiconductor chips contains 29 that are defective. Round your answers to four...

A lot of 106 semiconductor chips contains 29 that are defective. Round your answers to four decimal places (e.g. 98.7654).

a) Two are selected, at random, without replacement, from the lot. Determine the probability that the second chip selected is defective.

b) Three are selected, at random, without replacement, from the lot. Determine the probability that all are defective.

Expert Solution

SOLUTION:

From given data,

A lot of 106 semiconductor chips contains 29 that are defective.

Number of chips in a lot =106

Number of defective chips = 29

Suppose two chips are randomly selected without replacement.

Let E₁, denote selected chip Is defective in the first selection.

Let E₂, denote selected chip Is defective in the second selection.

Also,E^c₁ and E^c₂ are, compliments of E₁ and E₂ , respectively,

Probability of selecting a defective chip in the first selection , P( E₁) = 29 / 106

If the selected chip is defective in the first selection, then probability of selecting a defective chip in the second selection Is , P( E₂ IE₁ )=28 / 105,

If the selected chip is defective in the first selection, then probability of selecting a defective chip in the second selection Is , P( E₂ IE^c₁ )=29 / 105,

a) Two are selected, at random, without replacement, from the lot. Determine the probability that the second chip selected is defective.

Probability of selecting, a defective chip in the second selection,

P(E₂) = P(E₂ E₁ ) + P(E₂ E^c₁ )

= P(E₂ | E₁ ) P(E₁ )+ P(E₂ | E^c₁ ) P(E^c₁)

= (28 / 105)(29 / 106) + (29 / 105)(1-(29 / 106))

= (58 / 795) + (29 / 105)(77/106)

= (58/795)+(319/1590)

= 29/106

= 0.27358

Therefore, probability that the second chip selected is defective,
P(E₂)= 0.27358

b) Three are selected, at random, without replacement, from the lot. Determine the probability that all are defective.

Suppose three chips are randomly selected without replacement.

Compute the probability that all are defective.
P(all are defective) = ( 29 / 106) * (28/105)*(27/104)

= 0.0189

There fore , probability that all are defective is 0.0189

orchestra answered 2 years ago

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