Are My Calculations for Accident Recovery Probabilities Accurate?

AI Thread Summary
The discussion focuses on calculating probabilities related to accident recovery and the success of a new technique. For 20 accident victims with an 80% recovery rate, the probability of exactly 14 recovering is approximately 0.109, while the probability of at least 10 recovering is about 0.999. The probability of between 14 and 18 recoveries is around 0.849, and the chance of at most 16 recovering is about 0.589. In a separate scenario with a new technique performed five times, the probability of all five being successful at an 80% success rate is 0.32768, while the probability of at least three successes at a 90% success rate is 0.9271. These calculations provide insights into the effectiveness of recovery and technique success rates.
OptimusPrime
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Please focus on my numerical answers and let me know if they are correct.

Thanks

1. The probability that someone recovers from an accident is .8. Suppose 20 people are known to have been in accidents.

a. What is the probability that exactly 14 recover?

20 choose 14 * .8^14 * .2^6 =.109
b. What is the probability that at least 10 recover?

20 choose x * .8^x * .2^20-x
=1- P(Y<=9)= 1-.001 = .999

c. What is the probability that at least 14 but no more than 18 recover

20 choose x * .8^x * .2^20-x

P(14<=x<=18)= p(14) +p(15) + p(16) + p(17) + p(18) =.849

d.What is that probability that at most 16 recover?
20 choose x * .8^x * .2^20-x

P(x<=16)= p(0)+p(1)+p(2)+...p(16) = .589



2. A new technique has a probability of p. Assume the technique is peformed 5 times and the results are independent.
a. What is the prob that all 5 techniques are successful if p=.8?

5 choose 5 * .8^5 * .2^0 =.32768

b. What is the prob that exactly four are successful if p = .6?
5 choose 4 * .6^4 * .4^1 =.2592

c. What is the prob that less than two are successful if p=.3?

5 choose x * .3^x * .7^5-x

P(x<2) = p(0) + p(1) =.528
 
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d. What is the prob that at least three are successful if p=.9?
5 choose x * .9^x * .1^5-x

P(x>=3) = 1- P(x<=2) = 1-.0729 = .9271
 
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