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Take Test-Zachary Beadshaw Google Chrome Secure https //www.matholcom/Student/PIayerTestaspahes 21967 Zachary Bradshaw 328/172 26 BUSA 2100 B Spring 2017 Test: Midterm #2 Time Remaining: 00.57.17 Submit This Test: 93 pts pos This Question: 8 pts 7 of 1500 complete HE Question Help salesman normally makes a sale (closes) on 60% of his presentations, Assuming the presentations independent, sind the probability of each of the following a) He fails to close for the first time on his fourth attempt b He closes his first presentation on his third attempt c) The first presentation he closes wil be on his second attempt d The first presentation he closes wil be on one of his first three attempts a) The probability he fails to close for the fist time on his fourth attempt isU (Round to four decimal places b The probablity he closes his fist presentation on his thid amempt is (Round to four decimal places as needed c The probabany the finst presentation he closes wil be on his second anempt es as needed (Round to four decimal plac The probability the first presertation he coses wil be on one of his sist eree anemprs is (Round to four decimal places as needed in each of the answer boxesExplanation / Answer
a) Probability that he fails to close the deal first time on his fourth attempt= Probability that he closes the deal on all first 3 presentations * Probability that he could not close the deal on the 4th attempt.
=0.63*(1-0.6) = 0.216*0.4 = 0.0864
Therefore, 0.0864 is the required probability.
b) Probability that he closes his first presentation on 3rd attempt:
= (1-0.6)2*0.6 = 0.16*0.6 = 0.096
Therefore, 0.096 is the required probability.
c) Probability that he closes his first presentation on 2nd attempt:
= 0.4*0.6 = 0.24
Therefore, 0.24 is the required probability.
d) Probability that he closes his first presentation on one of his 3rd attempt:
= 0.6 + 0.4*0.6 + 0.4*0.4*0.6 = 0.6 + 0.24 + 0.096 = 0.936
Therefore, 0.936 is the required probability.
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