Probability-4 Easy one
A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target?
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3 solutions
solution - not my own.
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Ok, it can also be done by taking cases.
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Can you pls show how??
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@Ritu Roy – Take cases as 1- Probability that he hits the target once which means he fails in three attempts. 2-Probability that he hits the target twice and fails two times. .... ... Total 4 cases. Calculate the probability in each case and add them.
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@Mahimn Bhatt – @Mahimn Bhatt Thanx a lot
@Mahimn Bhatt – Is it like this
Pls correct if wrong
Case 1 P(Hits the target one time) = 4 1 ||y, Case 2 P(Hits the target two times) Case 3 P(Hits the target three times) Case 4 P(Hits the target four times) Total P = 4 1 + 2 1 + 4 3 + .... ???
Short and sweet
Efficient :)
Detailed solution. Upvoted.
A man can hit a target once in 4 shots. Obviously, he has a probability of 4 1 of hitting the target, and in 4 shots he has a probability of 4 1 ∗ 4 = 1 of hitting it. □
Well guess what? We could miss every single shot, couldn't he? So the probability can't be 1 . That's weird, isn't it?
No, not at all. We are effectively trying to find the probability that he'll hit the target at least once. Let's find the compliment: that he misses every single shot. This is obviously ( 4 3 ) 4 = 2 5 6 8 1 , so our probability is 1 − 2 5 6 8 1 = 2 5 6 1 7 5 . Or is it?
This can also be done using inclusion/exclusion. Try it yourself!
"A man can hit a target once in 4 shots" means that the probability of hitting the target is 1/4, thus the probability of missing the target is 1-1/4=3/4.
The probability that he will hit a target while shooting 4 times is basically the probability that he will hit at least once, so 1-(the probability that he will miss all 4 times): P=1- ( 4 3 ) 4 = 2 5 6 1 7 5