# The probability that three different archers (A, B, C) hit the mark, independently of one another, are respectively 1/6, 1/4, 1/3. Everyone shoots an arrow. Find the probability that only one hits the mark. If only one hits the mark, what is the probability that it is archer A?

To solve this problem, we only need to multiply the given probabilities with their complement.

So, for example, if the chance that archer A will hit a target is 1/6, then they have a 5/6 chance of not hitting the target. Clever manipulation of this information is enough.

The probability that only archer A will hit the target is given by:

P(A only) = (1/6) x (3/4) x (2/3) = 6/72

This is the answer to the second question. In the above case, notice that we take the complements of the probabilities given for archers B and C. This is because we're looking for the branches produced by archer A hitting, that result in B and C failing.

To get all the possibilities in which only a single archer hits, we have to perform a simple sum of the probabilities that each archer makes that lone hit. We already have archer A's.

This is for B:

P(B only) = (5/6) x (1/4) x (2/3) = 10/72

This is for C:

P(C only) = (5/6 x (3/4) x (1/3) = 15/72

Then, we can finally get the sum:

P(one hit) = P(A only) + P(B only) + P(C only)

P(one hit) = (6/72) + (10/72) + (15/72) = 31/72

Therefore, the probability that only a single arrow finds the mark is 31/72.

Supplementary calculations (feel free to ignore, just additional proof):

An indirect proof that we're using a correct method is to perform a simple sum with the remaining probabilities. Three uncomputed probabilities remain in this case, the probability that all archers hit, that only two archers hit, and no archers hit.

The probability that all archers hit is given by:

P(all) = (1/6) x (1/4) x (1/3) = 1/72

The probability that no archers hit is given by multiplying all the complements:

P(none) = (5/6) x (3/4) x (2/3) = 30/72

The probability that at least two is given by the sum of the following:

P(A & B) = (1/6) x (1/4) x (2/3) = 2/72

P(A & C) = (1/6) x (3/4) x (1/3) = 3/72

P(B & C) = (5/6) x (1/4) x (1/3) = 5/72

P(two hits) = P(A & B) + P(A & C) + P(B & C) = 10/72

The sum of all probabilities should add up to 72/72 or 1 or 100%.

P(total) = P(none) + P(one hit) + P(two hits) + P(all)

P(total) = (30/72) + (31/72) + (10/72) + (1/72) = 72/72 = 1