Question: In a shooting competition, probability of A hitting target is\(\frac{2}{5}\), by B is\(\frac{2}{3}\) and C is\(\frac{3}{5}\). If all of them fire independently, what is the probability that only one will hit target?
- \(\frac{3}{15}\)
- \(\frac{4}{25}\)
- \(\frac{1}{3}\)
- \(\frac{2}{3}\)
- \(\frac{3}{4}\)
“In a shooting competition, probability of A hitting target is\(\frac{2}{5}\), by B is\(\frac{2}{3}\) and C is\(\frac{3}{5}\).”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.
Solution and Explanation:
Approach Solution 1:
There is only one way to solve this problem, which is by using Probability.
Given:
- Probability of A hitting target is\(\frac{2}{5}\),
- B is \(\frac{2}{3}\)and
- C is \(\frac{3}{5}\)
Condition
- all of them fire independently
Find Out:
- What is the probability that only one will hit target
P(x) = probability of hitting target,
P(x") = probability of not hitting target = 1 - P(x), using this:
P(A) = \(\frac{2}{5}\), so, P(A") = \(\frac{3}{5}\)
P(B) =\(\frac{2}{3}\), so, P(B") = \(\frac{1}{3}\)
P(C) = \(\frac{3}{5}\), so, P(C") = \(\frac{2}{5}\)
OR = plus, and = *
We need: P(A hits & B doesnt & C doesnt) OR
P(A doesnt & B hits & C doesnt) OR
P(A doesnt & B doesnt & C hits)
which is: P(A)(B")(C") + P(A")P(B)P(C") + P(A")P(B")P(C)
= (⅖)*(⅓)*(⅖) + (⅗)*(⅔)*(⅖) + (⅓)*(⅗)
=\(\frac{(4+12+9)}{75}\)
=\(\frac{24}{75}\)
=\(\frac{1}{3}\)
Correct Answer: D
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