If the probability of hitting a target by a shooter, in any shot, is 1/3, then the minimum number of independent shots at the target required by him so that the probability of hitting the target at least once is greater than 5/6, is:

Let p be the probability of hitting the target in any shot. We are given that p = 1/3. Let n be the number of independent shots. The probability of not hitting the target in any shot is (1-p) = 1 - 1/3 = 2/3. The probability of not hitting the target in n shots is (2/3)^n. Therefore, the probability of hitting the target at least once in n shots is 1 - (2/3)^n. We are given that this probability must be greater than 5/6. So we have the inequality: 1 - (2/3)^n > 5/6. This simplifies to: (2/3)^n < 1 - 5/6 = 1/6. Taking the logarithm of both sides (with base 2/3): n > log(2/3)(1/6). Since log(2/3)(1/6) is approximately 3.7, n must be at least 4. To verify, let's check the probabilities for n=3 and n=4: For n=3: 1 - (2/3)^3 = 1 - 8/27 = 19/27 ≈ 0.7037 < 5/6 ≈ 0.8333 For n=4: 1 - (2/3)^4 = 1 - 16/81 = 65/81 ≈ 0.8025 < 5/6 ≈ 0.8333 For n=5: 1 - (2/3)^5 = 1 - 32/243 = 211/243 ≈ 0.8683 > 5/6 ≈ 0.8333 Therefore, the minimum number of shots required is 5.