Calculate binomial probabilities Binomial probability describes the likelihood of obtaining a specified number of successes in a fixed number of independent trials, each with two outcomes and the same success probability p p p . The calculation P ( X = k ) = ( n k ) p k ( 1 − p ) n − k P(X=k)=\binom{n}{k}p^k(1-p)^{n-k} P ( X = k ) = ( k n ) p k ( 1 − p ) n − k supports finding exact probabilities and ranges such as “at most” or “at least,” while interpreting n n n , p p p , and k k k ; the model does not apply when trials are dependent or their probabilities change. More advanced approximations and generalized distributions are beyond this scope.
Detailed Explanation: Calculate binomial probabilities A binomial probability is used when:
There is a fixed number of trials, n n n
Each trial has only two outcomes: success or failure.
The trials are independent.
The probability of success, p p p
The probability of getting exactly k k k
P ( X = k ) = ( n k ) p k ( 1 − p ) n − k , P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}, P ( X = k ) = ( k n ) p k ( 1 − p ) n − k , where:
n n n k k k p p p 1 − p 1-p 1 − p ( n k ) \binom{n}{k} ( k n ) k k k n n n
Example
A basketball player makes a free throw with probability 0.7 0.7 0.7 3 3 3 5 5 5
Step 1: Identify the values
There are 5 5 5 n = 5 n=5 n = 5
The desired number of successes is 3 3 3 k = 3 k=3 k = 3
The probability of success is p = 0.7 p=0.7 p = 0.7
The probability of failure is
1 − p = 1 − 0.7 = 0.3. 1-p=1-0.7=0.3. 1 − p = 1 − 0.7 = 0.3. Step 2: Substitute into the formula
P ( X = 3 ) = ( 5 3 ) ( 0.7 ) 3 ( 0.3 ) 5 − 3 P(X=3)=\binom{5}{3}(0.7)^3(0.3)^{5-3} P ( X = 3 ) = ( 3 5 ) ( 0.7 ) 3 ( 0.3 ) 5 − 3 Step 3: Calculate each part
( 5 3 ) = 10 \binom{5}{3}=10 ( 3 5 ) = 10 So,
P ( X = 3 ) = 10 ( 0.7 ) 3 ( 0.3 ) 2 P(X=3)=10(0.7)^3(0.3)^2 P ( X = 3 ) = 10 ( 0.7 ) 3 ( 0.3 ) 2 P ( X = 3 ) = 10 ( 0.343 ) ( 0.09 ) P(X=3)=10(0.343)(0.09) P ( X = 3 ) = 10 ( 0.343 ) ( 0.09 ) Step 4: Find the probability
P ( X = 3 ) = 0.3087 P(X=3)=0.3087 P ( X = 3 ) = 0.3087 Therefore, the probability that the player makes exactly 3 3 3 5 5 5
0.3087 \boxed{0.3087} 0.3087 or approximately 30.87 % \boxed{30.87\%} 30.87%