Hint for A. No. Draw a probability tree and look at the ways of getting a + result.
Hello everyone. I have encountered these problems. I tried to answer them, but I'm not sure if they are correct.:: I would be very grateful if someone will check my answers Thanks in advance...
A. A new diagnostic test has been developed fod mad cow disease (MCD).Among those animals with MCD, 90% were correctly identified by this new diagnostic test Also, the test has 95% success rate among those animals without MCD. Assuming that the probability that an animal has MCD is 0.05, what is the probability that an animal that tested positive by this diagnostic test has MCD?
Answer:
Let MCD : an animal has MCD
T: an animal is positive in MCD upon the test
P(T|MCD) = 0.90
P(MCD) = 0.05
P(T|MCDC ) = 0.95
P(MCD|T) = ? an animal tested positive by the diagnostic test has MCD
= PTMCDx PMCDPT⬚
= PTMCDx PMCDPT|MCD x PMCD+PTMCDCx P(MCDC)⬚
= (0.90)(0.05)0.900.05+0.950.95⬚⬚
= 0.0475 or 4.75%
B. Seventy percent (70%) of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that vehicles inspected successively pass or fail independently of one another, what is the probability that at least one of the next three vehicles fail inspection?
Let: P(pass) = 0.70
P(pass c) = 1 – P (pass) = 1 – 0.70 = 0.30
P(FAIL) = ½ = 0.50
Solution:
P(At least one of the next three vehicles fail) = (0.30) x (0.50)3
= 0.0375 or 3.75 %
C. A team was formed to investigate the proliferation of counterfeit coins. It was found that the weight of genuine coins is normally distributed with µ = 27 grams and = 4 grams, and the weight of counterfeit coins is normally distributed with µ = 23 grams and = 5 grams. It was also estimated that 80% of coins in circulation are genuine, and the rest are counterfeit.
1. What is the probability that a cashier gives you a counterfeit coin that weighs more than 25 grams?
P(G) = 0.80 genuine coins in the circulation
P(C) = 0.20 counterfeit coins in the circulation
P(C ≥ 25) = 25-235 = 2/5 or 0.40
= 1 – P(z ≤ 0.4) = 1 – 0.6554
= 0.3446
P(C ≥ 25) = (0.3446) x (0.20) = 0.0689 or 6.89% - probability that a cashier gives you a counterfeit coin that weighs more than 25 grams.
2. What is the probability that a randomly chosen genuine coin weighs more than 25 grams?
P(G ≥ 25) = 25-274 = - 2/4 or -0.50
= P(z ≤ 0.50) = 0.6915
P(G ≥ 25) = (0.6915) x (0.80) = 0.5532 or 55.32% - probability that a randomly chosen genuine coin weighs more than 25 grams.
3. What is the probability that the cashier will give you a coin that weighs more than 25 grams and is a counterfeit?
P(≥25 Ω C) = P(G ≥ 25) x P(C ≥ 25)
= 0.5532 x 0.0689
= 0.0381 or 3.81% - probability that a cashier will give you a coin that weighs more than 25 grams and is a counterfeit
4. What is the probability that the coin from the cashier that weighs more than 25 grams is counterfeit?
P(≥25|C) = P(≥25 Ω C)P(C)
= 0.03810.20 = 0.1905 or 19.05% - probability that the coin from the cashier that weighs more than 25 grams is counterfeit.
Thank you soo much for the help...
Hint for A. No. Draw a probability tree and look at the ways of getting a + result.
Hi. How about this answer for A.
P(A) = 0.90 correct identification of animals with MCD
P(B) = 0.95 success rate of diagnostic testin indentifying animals without MCD
P(C) = 0.05 animals with MCD
P(D) = animal tested positive by the diagnostic test has MCD
P(D) = P(A)×P(C)
= (0.90)×(0.05)
P(D) = 0.045 or 4.5%
hi,
just use the Bayes formula
P(MCD|T)=P(T|MCD)*P(MCD)/P(T)
where P(T)=P(T|MCD)*P(MCD)+P(T|MCDC)*P(MCDC)
regards
True, but it doesn't help us understand it much. Alternatively find the probability of a positive for both MCD and nonMCD, then take the required probability over the total.
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