Consider a binomial probability distribution with pequals040

Consider a binomial probability distribution with

pequals=0.40.4

and

nequals=99.

What is the probability of the following?

a)

exactly three successes

b)

less than three successes

c)

sevenseven

Solution

a)

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 =    9      
p = the probability of a success =    0.4      
x = the number of successes =    3      
          
Thus, the probability is          
          
P (    3   ) =    0.250822656 [ANSWER]

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b)

Note that P(fewer than x) = P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    9      
p = the probability of a success =    0.4      
x = our critical value of successes =    3      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   2   ) =    0.231787008
          
Which is also          
          
P(fewer than   3   ) =    0.231787008 [ANSWER]

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c)

Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    9      
p = the probability of a success =    0.4      
x = our critical value of successes =    7      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   6   ) =    0.974965248
          
Thus, the probability of at least   7   successes is  
          
P(at least   7   ) =    0.025034752 [ANSWER]