Question 483897: the probability that a bullet fired from a point will hit the target is 1/3. Three such bullets are fired simultaneously towards the target from that very point. What is the probability that the target will be hit?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! p = probability that the target will be hit = 1/3
q = probability that the target will not be hit = 1 - 1/3 = 2/3
p + q = 1, as it should, since either the target will be hit or the target will not be hit.
the probability that the target will be hit is equal to 1 - the probability that the target will not be hit.
the probability that the target will not be hit is 2/3 * 2/3 * 2/3 because you are firing at the target 3 times and each time the probability that the target will not be hit is 2/3.
the probability that the target will not be hit is therefore 8/27 (2*2*2)/(3*3*3).
since the total probability is 1 which is equivalent to 27/27, then the probability that the target will be hit is equal to 27/27 - 8/27 = 19/27
this is the short way to solve this problem.
the long way is to look at the probability that:
a. the target is hit 1 time only.
b. the target is hit 2 times only.
c. the target is hit 3 times only.
since the target can't be hit any other way, then the probability that the target will be hit will be the sum of these 3 independent events.
the probability that any of the posssible events is given by the formula:
nCx * p^x * q^(n-x)
n is the total number of shots fired at the target.
x is the number of times the target can be hit exactly.
this comes out to be:
3C0 * p^0 * q^3
3C1 * p^1 * q^2
3C2 * p^2 * q^1
3C3 * p^3 * q^0
We already solved for 3C0 * p^0 * q^3
that number came out to be 8/27
3C0 equals 3!/(0!*3!) which equals 1
p^0 = 1
q^3 = (2/3) * (2/3) * (2/3) which equals 8/27
3C0 * p^0 * q^3 = 1 * 1 * 8/27 = 8/27
We can solve for 3C3 * p^3 * q^0 easy enough.
3C3 equals (3! / (3!*0!) which equals 1
p^3 = (1/3) * (1/3) * (1/3) which equals 1/27
q^0 = 1
3C3 * p^3 * q^0 = 1 * 1/27 * 1 = 1/27
We can solve for 3C1 * p^1 * q^2 as follows:
3C1 = 3! / (1!*2!) = 3!/2 = 6/2 = 3
p^1 = 1/3
q^2 = 2/3 * 2/3 = 4/9
3C1 * p^1 * q^2 = 3 * 1/3 * 4/9 = 3/3 * 4/9 = 12/27
We can solve for 3C2 * p^2 * q^1 as follows:
3C2 = 3! / (2!*1!) = 3!/2 = 6/2 = 3
p^2 = 1/3 * 1/3 = 1/9
q^1 = 2/3
3C2 * p^2 * q^1 = 3 * 1/9 * 2/3 = 3/9 * 2/3 = 6/27
our total probabilities are:
p(0) = 8/27
p(1) = 12/27
p(2) = 6/27
p(3) = 1/27
The total probability is 8 + 12 + 6 + 1 = 27/27 = 1 as it should.
The probability we will hit the target is p(1) + p(2) + p(3) = 12 + 6 + 1 = 19/27.
That's the same number we got before when we did 1 - p(0).
We're good and that's your answer.
The probability that the target will be hit is 19/27.
nCx is the combination formula and it is defined as n! / (x! * (n-x)!)
You need to multiply the probability of each occurrence by the number of ways that occurrence can occur.
An example:
You have 3 shots.
You want to get 1 out of 3 shots.
The probability is 3C1 * p(1) * q(2).
Assume h means hit and x means miss.
The number of ways is 3C2 which is equal to 3 (we did that above).
Those possible ways are:
hmm
mhm
mmh
h = hit
m = miss
The probability of each of these ways is the same.
The probability of getting 1 out of 3 is p(hmm) + (pmhm) + p(mmh).
Since the probability of hmm is equal to the probability of mhm is equal to the probability of mmh, then your probability of getting 1 out of 3 can be expressed as 3 * p(hmm).
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