The probability of a man hitting a target is 1/10. The least number of shots required, so that the probability of his hitting the target at least once is greater than 1/4, is ____________.

Solution

Maths sir · 3 steps

For 'at least once', work with the complement: find when missing every shot becomes unlikely enough.

1. Use the complement. Each shot misses with probability 1 - 1/10 = 9/10, and the shots are independent. So the chance of missing all n shots is (9/10)ⁿ, and the chance of hitting at least once is 1 - (9/10)ⁿ.

   1 - (9/10)ⁿ > 1/4 ⇒ (0.9)ⁿ < 3/4 = 0.75

2. Try small n. n = 1 gives 0.9 and n = 2 gives 0.81. Both are more than 0.75, so they fail. n = 3 gives 0.729, which is less than 0.75.
3. So. With 3 shots the probability of at least one hit is 1 - 0.729 = 0.271 > 0.25, and 2 shots are not enough. The least number of shots is 3.

The idea to keep. Whenever you see 'at least once', write it as 1 minus the probability of 'never'. Then the condition becomes a power of the failure probability, and you can test small values of n.

Answer from the NTA final answer key. Solution by PrepWiser.