Question

Hanz and Franz are in a shooting competition. The object of the match is to be the first to hit the bull's eye of a target 100 feet away. The two opponents alternate turns shooting, and each opponent has a 40% chance of hitting the bull's-eye on a given shot. If Hans graciously allows Franz to shoot first, what is the probability that Hans will win the competition and take no more than three shots?

Answer 1

Answer: The probability of Hans winning and taking no more than three shots can be calculated using the principle of total probability and the law of total probability.

Let's call the event of Hans winning and taking no more than three shots "A". Then, the probability of event A can be broken down into several cases:

Case 1: Hans wins in one shot: P(A|Hans wins in one shot) = P(Hans wins in one shot) = 0.4.

Case 2: Hans wins in two shots: P(A|Hans wins in two shots) = P(Franz misses in the first shot) * P(Hans wins in the second shot) = (1 - 0.4) * 0.4 = 0.24.

Case 3: Hans wins in three shots: P(A|Hans wins in three shots) = P(Franz misses in the first two shots) * P(Hans wins in the third shot) = (1 - 0.4)^2 * 0.4 = 0.144.

Finally, the probability of event A can be calculated using the law of total probability as follows:

P(A) = P(A|Hans wins in one shot) + P(A|Hans wins in two shots) + P(A|Hans wins in three shots) = 0.4 + 0.24 + 0.144 = 0.784.

So, the probability of Hans winning the competition and taking no more than three shots is 0.784.

Explanation: