Need Help please and Thanks ALGEBRA 2Alternate Exam 6continu
Solution
For multiple, or at least 2 Jennifer\'s:
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 =    20      
 p = the probability of a success =    0.0403      
 x = our critical value of successes =    2      
           
 Then the cumulative probability of P(at most x - 1) from a table/technology is          
           
 P(at most   1   ) =    0.955313118
           
 Thus, the probability of at least   2   successes is  
           
 P(at least   2 Jennifer\'s) =    0.044686882
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For multiple, or at least 2 Emily\'s:
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 =    20      
 p = the probability of a success =    0.0136      
 x = our critical value of successes =    2      
           
 Then the cumulative probability of P(at most x - 1) from a table/technology is          
           
 P(at most   1   ) =    0.997588701
           
 Thus, the probability of at least   2   successes is  
           
 P(at least   2 Emily\'s   ) =    0.002411299
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Thus, the ratio of probabilities is
P(at least 2 Jennifers)/P(at lest 2 Emilys) = 18.53 [ANSWER]
Thus, it is 18.53 times more likely to have multiple Jennifers in a group of 20 than multiple Emilys in a group of 20.

