Question

Carl hits the target 50% pf the time he throws a ball at it. Carl uses a coin to simulate his next three pitches. The assigns H for a hit and T for a miss. The results of the 12 trials are shown below. Based on the results, what is the probability that cael will hit the next the Target with exactly two of his next three throws

Answer 1

The probability that Carl will hit the target with exactly two of his next three throws is 37.5%.

Step-by-step explanation:

To find the probability that Carl will hit the target with exactly two of his next three throws, we need to determine the number of favorable outcomes and the total number of possible outcomes. From the given information, we know that Carl hits the target 50% of the time, which means he misses the target the other 50% of the time.

In this case, we can use the binomial probability formula to calculate the probability:

P(X=k) = C(n, k) * p^k * (1-p)^(n-k)

Where:

• P(X=k) is the probability of getting k successes (hitting the target) out of n trials (throws)
• C(n, k) is the number of combinations of n items taken k at a time
• p is the probability of success (hitting the target)
• (1-p) is the probability of failure (missing the target)

In this case, n=3 (the number of throws) and p=0.5 (the probability of hitting the target).

Let's calculate the probability:

P(X=2) = C(3, 2) * (0.5^2) * (1-0.5)^(3-2) = 3 * 0.25 * 0.5 = 0.375

Therefore, the probability that Carl will hit the target with exactly two of his next three throws is 0.375, or 37.5%.