A restaurant serves 8 fish entrees 12 beef and 10 chicken If

A restaurant serves 8 fish entrees, 12 beef, and 10 chicken. If customers select from these entrees randomly, what is the probability:

(a) that exactly 2 of the next 4 customers order fish?

(b) that more than half of the next 8 customers order fish?

(b) that of 20 customers, none order chicken?

Solution

a)

Here, P(fish) = 8/30 = 0.2666667.

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 =    4      
p = the probability of a success =    0.266666667      
x = the number of successes =    2      
          
Thus, the probability is          
          
P (    2   ) =    0.229451852 [ANSWER]

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b)

This happens when at least 5 order fish.

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 =    8      
p = the probability of a success =    0.266666667      
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   ) =    0.964216405
          
Thus, the probability of at least   5   successes is  
          
P(at least   5   ) =    0.035783595 [ANSWER]

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c)

Here, P(chicken) = 10/30 = 0.33333333.

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 =    20      
p = the probability of a success =    0.333333333      
x = the number of successes =    0      
          
Thus, the probability is          
          
P (    0   ) =    0.000300729 [ANSWER]