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LAB 26 (8.3) MATH 1319 SHOW ALL WORK 5. If a salesperson has gross sales of over

ID: 3316772 • Letter: L

Question

LAB 26 (8.3) MATH 1319 SHOW ALL WORK 5. If a salesperson has gross sales of over $600,000 in a year, then he or she is cigible to play the company's bonus game: A black box contains 2 one-dollar bills, 1 five-dolar bill and 1 twenty-dollar bil Bis are drawn out of the box one at a time without replacement unbl a twenty-dollar bill is drawn. Then the game stops. The salesperson's bomas is 1,000 times the value of the biüls drawn. Complete parts (A) through (C) below. A. What is the probablity of winning a $26,000 bonus? 1O 55 I 16 45 45 24 What is the probablity of winming the maximum bomas by drawing ou al the bill som the box? B. C. What is the probabdlity of the game stopping at the third draw?

Explanation / Answer

A.

The salesperson will receive $26000 bonus if he/she draws 1 one-dollar bill, 1 five-dollar bill and at last 1 twenty-dollar bill. (to get total $26 bills)
The possible draws to get this combination is,
(1,5,20)
or, (5,1,20)

P(1,5,20) = (2/4) * (1/3) * (1/2) = 1/12
P(5,1,20) = (1/4) * (2/3) * (1/2) = 1/12

So, the probability that salesperson will receive $26000 bonus = (1/12) + (1/12) = 1/6

B.
To get the maximum bonus, the salesperson should draw the twenty dollar bill at the last.
Total number of bills other than twenty dollar are 2 + 1 = 3
Probability of maximum bonus = Probability to get one or two dollar bills in 3 draws = (3/4) * (2/3) * (1/2) = 1/4

C.
The game will stop in 3rd draw, if we get the 20 dollar bill in 3rd draw.
So, the salesperson would draw one or two dollar bill in first 2 draws.
The possible draws to get this combination is,
(1,5,20)
or, (5,1,20)
or, (1,1,20)

P(1,5,20) = (2/4) * (1/3) * (1/2) = 1/12
P(5,1,20) = (1/4) * (2/3) * (1/2) = 1/12
P(1,1,20) = (2/4) * (1/3) * (1/2) = 1/12
So, the probability that game will stop in 3rd draw = (1/12) + (1/12) + (1/12) = 1/4