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.
