Suppose 93 of all students taking a beginning programming co

Suppose 93% of all students taking a beginning programming course fail to get their first program to run on the first submission. We are interested in probabilities for a group of 6 such students. This can be considered a binomial distribution.

Among the 6 students, find the probability that:

e. All fail on their first submissions

f. At least 5 fail on their first submission

g. Less than 5 fail on their first submission

h. What is the mean number who will fail?

Solution

E)

P(all fail) = 0.93^6 = 0.646990183 [answer]

F)

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    6      
p = the probability of a success =    0.93      
x = our critical value of successes =    5      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   4   ) =    0.060820701
          
Thus, the probability of at least   5   successes is  
          
P(at least   5   ) =    0.939179299 [answer]

g)

Note that P(fewer than x) = P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    6      
p = the probability of a success =    0.93      
x = our critical value of successes =    5      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   4   ) =    0.060820701
          
Which is also          
          
P(fewer than   5   ) =    0.060820701 [answer]

h)

mean = n p = 6*0.93 = 5.58 [ANSWER]