Let the probability of success on a Bernoulli trial be 028 I

Let the probability of success on a Bernoulli trial be 0.28.

In five Bernoulli trials, what is the probability that there will be 4 failures? (Round your intermediate calculations and final answers to 4 decimal places.)

In five Bernoulli trials, what is the probability that there will be more than the expected number of failures? (Round your intermediate calculations and final answers to 4 decimal places.)

Solution

A)

If there are 4 failures, then there is 1 success.

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    5      
p = the probability of a success =    0.28      
x = the number of successes =    1      
          
Thus, the probability is          
          
P (    1   ) =    0.376233984 [ANSWER]

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b)

The expected number of failures is

n(1-p) = 5*(1-0.28) = 3.6.

Thus, there must be at least 4 failures, or, at most 1 success.

Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    5      
p = the probability of a success =    0.28      
x = the maximum number of successes =    1      
          
Then the cumulative probability is          
          
P(at most   1   ) =    0.569725747 [ANSWER]