bernoulli trial

Question

bernouilli trial with probability of success p=.33 is performed 7 times. find the probability of getting at most 5 successes. 

Answer 1

The probability of at most 5 successes is denoted as:

 

    P(0 <= X <= 5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)

 

Note:

 

The equation for a Bernoulli trial is: P(k successes in ntrials) = (n choose k)(p^k)(q^n-k)

 

(n choose k) is solved using:  n!/k!(n-k)!  as long as k < n

n = number of trials
k = number of successes
n – k = number of failures
p = probability of success in one trial
q = 1 – p = probability of failure in one trial

 

so, let's fill in the letters

 

n = 7

k = 0,1,2,3,4,5

n-k = 7,6,5,4,3,2

p = 0.33

q = 1-p = 1-0.33 = 0.67

 

Now solve for: n choose k

 

7 choose 0 = 7!/0!(7-0)! = 7!/0! 7! = 1

7 choose 1 = 7!/1!(7-1)! = 7!/1! 6! = 7

7 choose 2 = 7!/2!(7-2)! = 7!/2! 5! = 21

7 choose 3 = 7!/3!(7-3)! = 7!/3! 4! = 35

7 choose 4 = 7!/4!(7-4)! = 7!/4! 3! = 35 

7 choose 5 = 7!/5!(7-5)! = 7!/5! 2! = 21

 

Now we take the equation for a Bernoulli trial [(n choose k)(p^k)(q^n-k)] and set one equation up for each number that k represents and solve.

 

P(X=0) = (1)(0.33⁰)(0.67^7-0) = 0.060607116

P(X=1) = (7)(0.33¹)(0.67^7-1) = 0.208958863

P(X=2) = (21)(0.33²)(0.67^7-2) = 0.308760111

P(X=3) = (35)(0.33³)(0.67^7-3) = 0.253459792

P(X=4) = (35)(0.33⁴)(0.67^7-4) = 0.124838405

P(X=5) = (21)(0.33⁵)(0.67^7-5) = 0.036892544

 

Now add the results together as shown in the first equation.

 

P(0 <= X <= 5) = 0.060607116 + 0.208958863 + 0.308760111 + 0.253459792 + 0.124838405 + 0.036892544

                       = 0.993516831