Binomial probability Q's (1 Viewer)

[5647382910]

These are from Fitzpatrick ex. 29(b) q's 20,23,27 respectively:

20) On the average, a typist has to corect one word in 800. Assuming that a page contains 200 words, find the probability of more than one correction per page. ANSWER = 0.0625

23) A marksman finds that he hits the target 9 out of every 10 times and scores a bullseye once every 5 rounds. He fires 4 rounds. What is the probability that he scores at least 2 bullseyes and he has hit the target on each of the 4 rounds. ANSWER = 177/1250

27) A factory has 7 machines, 4 of model A which are in use, on the average, 80% of the timeand 3 of odel B which are in use, on the average, 60% of the time. If the foreman walks into the factory at a randomly selected time, what is the probability that he will find 2 machines of model A and 1 of model B in use. ANSWER = 0.04422

Any help would be much appreciated.

 

[5647382910]

Help...anyone?

 

[Angel_a]

q.20)
P(1 correction) = 1/800
P (no correction) = 799/800

P(more than 1 correction)
= 1-P(no correction) - P(1 correction)

now P(no correction) = (799/800)^200
P (1 correction) = (1/800) (799/800)^199 x 200C1

therefore
P (more than 1 correction)
= 1- (799/800)^200 - (1/800) (799/800)^199 x 200C1
= 0.0264

q.23)
P(at least 2 eye + 4 targets)

Now, the probability of getting the bull's eye is 2/10
the probability of getting the target is 9/10
ie the probability of getting the target but not the bull's eye is 7/10

therefore
P(getting all 4 bulls eye) = (1/5)^4
P (3 eye 1 target) = (1/5)^3 x (7/10) x 4C3
P (2 eye 2 target) = (1/5)^2 x (7/10)^2 x 4C2

add all together = 177/1250


im still doing the exercise.. haven't up to q.27 yet..
btw.. do u know how to do q.26?
and my answers for q.24 are a bit weird too..