An NBA player makes a successful free-throw 80% of the time. Find the probability that he is successful at least 4 times in 5 attempts?
we want the prob of
4 out of 5 + 5 out of 5
= C(5,4)(.8^4)(.2) + C(5,5) (.8^5)
= .4096 + .32768
= .73728
To find the probability that the player is successful at least 4 times in 5 attempts, we can use the binomial probability formula.
The probability of success on each attempt is 80% or 0.8 (because the player makes a successful free-throw 80% of the time).
Let's calculate the probability of getting exactly 4 successful attempts in 5 attempts and the probability of getting exactly 5 successful attempts in 5 attempts. Then we can add these two probabilities together to get the final result.
Probability of getting exactly 4 successful attempts in 5 attempts:
P(X = 4) = C(5, 4) * (0.8)^4 * (1-0.8)^1
= 5 * 0.8^4 * 0.2
Probability of getting exactly 5 successful attempts in 5 attempts:
P(X = 5) = C(5, 5) * (0.8)^5 * (1-0.8)^0
= 1 * 0.8^5 * 1
Now let's plug in the values and calculate:
P(X = 4) = 5 * 0.8^4 * 0.2
P(X = 5) = 1 * 0.8^5 * 1
Adding these probabilities together:
P(X >= 4) = P(X = 4) + P(X = 5)
So, the probability that the player is successful at least 4 times in 5 attempts is P(X >= 4).