SUPERCOMP, a retail computer store, sells personal computers and printers. The n
ID: 1191268 • Letter: S
Question
SUPERCOMP, a retail computer store, sells personal computers and printers. The number of computers (y) and printers (x) sold on any given day varies, with the probabilities of various possible sales outcomes being given by the following table:
-Number of Computers sold (y) [horizontal set]
-Number of printer sold (x) [vertical set]
a. What is the probability that 1 printer and 1 computer are sold?
b. What is the probability that 0 printers are sold?
c. What is the probability that 0 computers are sold?
d. What is the probability that at least 2 computers are sold?
e. What is the probability that at least 2 printers are sold?
f. What is the probability 3 printers are sold, given the information that 3
computers are sold?
g. What is the probability that 3 printers are sold, given that atleast 3 computers
are sold?
Explanation / Answer
(a) Probability of 1 printer & 1 computer sold = 0.05 = 5%
(b) Probability of 0 printer sold = 0.03 + 0.02 + 0.01 + 0.01 + 0.01 = 0.08 = 8%
(c) Probability of 0 computer sold = 0.03 + 0.03 + 0.02 + 0.02 + 0.01 = 0.11 = 11%
(d) Probability of at least 2 computers sold = P(2 computers) + P(3 computers) + P(4 computers)
= (0.01 + 0.02 + 0.1 + 0.05 + 0.05) + (0.01 + 0.01 + 0.05 + 0.1 + 0.1) + (0.01 + 0.01 + 0.05 + 0.1 + 0.16)
= 0.83
(e) Probability of at least 2 printers sold = P(2 printers) + P(3 printers) + P(4 printers)
= (0.02 + 0.06 + 0.1 + 0.05 + 0.01) + (0.02 + 0.02 + 0.05 + 0.1 + 0.01) + (0.01 + 0.01 + 0.05 + 0.1 + 0.16)
= 077
(f) P(3 Printers | 3 computers) = 0.10 [Row-column intersection cell of 3 printers, 3 computers]
(g) P(3 printers | at least 3 computers) = 0.1 + 0.05 = 0.15
NOTE: The table provided was not confusing at all :)
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