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)=\\\\\n20\/100\\cdot 19\/99+80\/100\\cdot20\/99=\\\\\n1\/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)=\\\\\nP(A_3|A_1A_2)P(A_1A_2)+\\\\\nP(A_3|A_1A_2^c)P(A_1A_2^c)+\\\\\nP(A_3|A_1^cA_2)P(A_1^cA_2)+\\\\\nP(A_3|A_1^cA_2^c)P(A_1^cA_2^c)=\\\\"
"18\/98\\cdot19\/99\\cdot20\/100+\\\\\n19\/98\\cdot80\/99\\cdot20\/100+\\\\\n19\/98\\cdot20\/99\\cdot80\/100+\\\\\n20\/98\\cdot79\/99\\cdot80\/100="
"18\/98\\cdot19\/99\\cdot1\/5+\\\\\n19\/98\\cdot80\/99\\cdot1\/5+\\\\\n19\/98\\cdot80\/99\\cdot1\/5+\\\\\n20\/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.
Comments
Leave a comment