Probability Question?

Question

A medical screening program administers three independent fitness tests. Of the persons taking the tests, 80% pass test I, 75% pass test II, and 60% pass test III. A participant is chosen at random
What is the probability that she will pass at least 2 of the 3 tests?

A tv set contains five circuit boards of type A, five of type B, and three of type C. The probability of failing in its first 50000 hours of use is .01 for type A, .02 for type B, and .025 for type C. Assuming that the failures of the various circuit boards are independent of one another, compute the probability that no circuit board fails in the first 5000 hours of use.

The pobability that a person A and a person B will live an additional 15 years are .8 and .7 respectively. Assuming that their lifespans are independent, what is the probability that A or B will live an additional 15 years?

If you know how to solve please show your work so that I can learn how to do it THANK YOU

Ooops in question 2 both should be 5000 not 50000 sorry

Answer 1

For the first one, there are 4 cases you must consider. 3 cases in which the person passes two and fails one, and 1 case where he passes all three.

The total prob will be
0.8*0.75*(1-0.6)
+ 0.8*(1-0.75)*0.6
+ (1-0.8)*(0.75)*(0.6)
+ 0.8*0.75*0.6

Add that up.

Answer 2

1. The med tests. This is a "tree" analysis type.
Let A, B, and C be the 3 tests.
Pass A, Fail B, Pass C = .8x.25x.6=.12
Pass A, Pass B = .8x.75= .60
(we dont care if they pass C or not)
Fail A, Pass B, Pass C = .20x.75x.6= .09
Total = 0.81
3. This is similar, since only one has to live for this probability. Failure is that neither do, so
p(neither live 15 years)= .2 x .3 =0.06
2. I don't know if you have a typo with the time to failure.

Answer 3

1. 80 x 0.75= 60%

2. type A = 100 - 0.01/(50000/5000)/5= 99.9998%
type B = 100 - 0.02/(50000/5000)/5
type B = 100 - 0.025/(50000/5000)/3

3. 0.8+0.7 = 1.5.. ( not sure about this one)

Answer 4

I didn't understand ur question.
Anyway thanks for giving me 2 points.