An NBA player makes a successful free-throw 80% of the time. Find the probability that he is successful exactly 4 times in 5 attempts?
Binomial probability function:
P(x) = (nCx)(p^x)[q^(n-x)]
x = 4
n = 5
p = .8
q = 1 - p = .2
With your data:
P(4) = (5C4)(.8^4)[.2^(5-4)]
I'll let you finish the calculation.
To find the probability that an NBA player makes a successful free-throw exactly 4 times in 5 attempts, we can use the binomial probability formula. The binomial probability formula is given by:
P(x) = nCx * p^x * (1-p)^(n-x)
Where:
P(x) = probability of exactly x successes
n = number of trials
x = number of successful outcomes
p = probability of a successful outcome in a single trial (in this case, making a successful free-throw)
In this case, the player attempts a free-throw 5 times, and has an 80% chance of making a successful free-throw in a single attempt (p = 0.8).
Using the formula, we can substitute the values:
P(4) = 5C4 * 0.8^4 * (1-0.8)^(5-4)
Let's calculate each component:
5C4 = (5! / (4! * (5-4)!)) = 5 (since 5! is equal to 5 multiplied by 4!)
0.8^4 = 0.4096 (since 0.8 * 0.8 * 0.8 * 0.8 = 0.4096)
(1-0.8)^(5-4) = (0.2)^1 = 0.2 (since 1-0.8 is equal to 0.2)
Now we can substitute these values into the formula:
P(4) = 5 * 0.4096 * 0.2
P(4) = 0.4096
Therefore, the probability that the NBA player makes a successful free-throw exactly 4 times in 5 attempts is 0.4096 or 40.96%.