A basket ball player makes 70% of her free throws. She takes 7 free throws in a game, and makes 5 of her 7 throws. If the shots are independent of each other, Then what is the probability that she makes the first five and misses the last 2?

To find the probability that the basketball player makes the first five free throws and misses the last two, we can use the concept of independent events.

Let's break down the problem step by step:

Step 1: Determine the probability of making a free throw.
The basketball player makes 70% of her free throws, so the probability of making a single free throw is 0.7.

Step 2: Calculate the probability of missing a free throw.
Since the probability of making a free throw is 0.7, the probability of missing a free throw is 1 - 0.7 = 0.3.

Step 3: Analyze the sequence of events.
We want to find the probability of making the first five free throws and missing the last two. Since each free throw is independent, we can simply multiply the probabilities together.

Step 4: Calculate the probability using multiplication.
P(making first five and missing last two) = P(making the first free throw) * P(making the second free throw) * ... * P(making the fifth free throw) * P(missing the sixth free throw) * P(missing the seventh free throw)

P(making first five and missing last two) = 0.7 * 0.7 * 0.7 * 0.7 * 0.7 * 0.3 * 0.3

Step 5: Perform the calculation.
P(making first five and missing last two) = 0.7^5 * 0.3^2

Step 6: Calculate the final result.
P(making first five and missing last two) = 0.16807

Therefore, the probability that she makes the first five free throws and misses the last two is approximately 0.16807, or 16.807%.

To find the probability that the basketball player makes the first five free throws and misses the last two, we need to multiply the individual probabilities of each event happening.

Given that the basketball player makes 70% of her free throws, the probability of making a free throw is 0.7, and the probability of missing a free throw is 0.3.

Since the shots are independent of each other, the probability of making the first five free throws is:

P(make) * P(make) * P(make) * P(make) * P(make) = 0.7 * 0.7 * 0.7 * 0.7 * 0.7 = 0.7^5 ≈ 0.16807

Similarly, the probability of missing the last two free throws is:

P(miss) * P(miss) = 0.3 * 0.3 = 0.3^2 = 0.09

To find the probability of making the first five and missing the last two, we multiply the two probabilities:

P(make first five) * P(miss last two) = 0.16807 * 0.09 ≈ 0.01517

Therefore, the probability that the basketball player makes the first five free throws and then misses the last two is approximately 0.01517 or 1.517%.

You want the specific case of

HHHHHMM, where H is success and M is a miss.

prob(HHHHHMM) = (.7)^5 (.3)^2
= ....