Question

A lot of 100 semiconductor chips contains 20 that are defective. (a) Two chips are selected at random, without replacement, from the lot. Use the total probability rule to determine the probability that the second chip selected is defective. (b) Three chips are selected at random, without replacement, from the lot. Use the total probability rule to determine the probability that the third chip selected is defective.

Answer 1

Define events

A₁ = first selected chip is defective,

A₂ = second selected chip is defective

A₃ = third selected chip is defective.

a)

Conditional Probabilities

$P(A_2|A_1)=\frac{19}{99}$

$P(A_2|A_1^c)=\frac{20}{99}$

$A_1$ and $A_1^c$ form complete set of events,

using the total probability rule

$P(A_2)=P(A_1)P(A_2|A_1)+P(A_1^c)P(A_2|A_1^c)=\\ 20/100\cdot 19/99+80/100\cdot20/99=\\ 1/5\cdot19/99+4/5\cdot20/99=1/5\cdot(19+80)/99=1/5$

b)

$P(A_1A_2)=P(A_2|A_1)P(A_1)=19/99\cdot20/100$

$P(A_1A_2^c)=P(A_2^c|A_1)P(A_1)=80/99\cdot20/100$

$P(A_1^cA_2)=P(A_2|A_1^c)P(A_1^c)=20/99\cdot80/100$

$P(A_1^cA_2^c)=P(A_2^c|A_1^c)P(A_1^c)=79/99\cdot80/100$

$A_1A_2, A_1A_2^c,A_1^cA_2,A_1^cA_2^c$ form complete set of events.

$P(A_3)=\\ P(A_3|A_1A_2)P(A_1A_2)+\\ P(A_3|A_1A_2^c)P(A_1A_2^c)+\\ P(A_3|A_1^cA_2)P(A_1^cA_2)+\\ P(A_3|A_1^cA_2^c)P(A_1^cA_2^c)=\\$

$18/98\cdot19/99\cdot20/100+\\ 19/98\cdot80/99\cdot20/100+\\ 19/98\cdot20/99\cdot80/100+\\ 20/98\cdot79/99\cdot80/100=$

$18/98\cdot19/99\cdot1/5+\\ 19/98\cdot80/99\cdot1/5+\\ 19/98\cdot80/99\cdot1/5+\\ 20/98\cdot79/99\cdot4/5=$

$\frac{18\cdot19+19\cdot80+19\cdot80+20\cdot79\cdot4}{98\cdot99\cdot5}=\cfrac{9702}{9702\cdot5}=\frac{1}{5}$

Answer: a) 1/5, b) 1/5.