Are My Calculations for Accident Recovery Probabilities Accurate?

[OptimusPrime]

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

 

[anathema]

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) = p(3) + p(4) + p(5) = .99701

 

[wais]

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