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wrote...
Posts: 221
2 weeks ago
A survey of local car dealers revealed that 64% of all cars sold last month had CD players, 28% had alarm systems, and 22% had both CD players and alarm systems.

a. What is the probability one of these cars selected at random had neither a CD player nor an alarm system?
b. What is the probability that a car had a CD player unprotected by an alarm system?
c. What is the probability a car with an alarm system had a CD player?
d. Are having a CD player and an alarm system disjoint events? Explain.
Textbook 

Stats: Modeling the World


Edition: 4th
Authors:
Read 41 times
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wrote...
2 weeks ago
a.  P(CD ∪ A) = P(CD) + P(A) - P(CD ∩ A) = 0.64 + 0.28 - 0.22 = 0.70,
     so P(neither) = 1 - 0.70 = 0.30
Or, using the Venn diagram below, 30%

b.  P(CD ∩ AC) = 0.64 - 0.22 = 0.42
c.  P(CD) = = ≈ 0.786
d.  No. P(CD ∩ A) = 0.22.   If they were disjoint, the probability of a car having both features would be 0.
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