kbuhvu
19.01.2021 •
Mathematics
A (blindfolded) marksman finds that on the average he
hits the target 4 times out of 5. If he fires 4 shots, what
is the probability of :
(a) more than 2 hits?
(b) at least 3 misses?
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Ответ:
Using the binomial distribution, it is found that there is a:
a) 0.9728 = 97.28% probability of more than 2 hits.
b) 0.0272 = 2.72% probability of at least 3 misses.
For each shot, there are only two possible outcomes, either he hits the target, or he does not. The probability of hitting the target on a shot is independent of any other shot, hence, the binomial distribution is used to solve this question.
Binomial probability distribution
The parameters are:
x is the number of successes. n is the number of trials. p is the probability of a success on a single trial.In this problem:
Hits the target 4 times out of 5, hence .He fires 4 shots, hence .Item a:
This probability is:
Then:
0.9728 = 97.28% probability of more than 2 hits.
Item b:
At least 3 misses is less than 2 hits, hence:
0.0272 = 2.72% probability of at least 3 misses.
To learn more about the binomial distribution, you can take a look at link
Ответ:
lo siento pero a eso no le entiendo