# Answer in Statistics and Probability for hiiie #242092

Consider randomly selecting a single individual and having that person test

drive 3 different vehicles. Define events A1, A2, and A3 by A1=likes vehicle #1,

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Answer in Statistics and Probability for hiiie #242092
Just from $13/Page A2 =likes vehicle #2, A3 =likes vehicle #3. Suppose that ( ) 65.0 P A1 = , ( ) 55.0 P A2 = , ( ) 70.0 P A3 = , ( ) 80.0 P A1 ∪ A2 = , ( ) ,60.0 P A1 ∪ A3 = ( ) 40.0 P A2 ∪ A3 = , ( ) 88.0 P A1 ∪ A2 ∪ A3 = . a. What is the probability that the individual likes both vehicle #1 and vehicle #3? b. Determine and interpret ( / ) P A1 A3. c. Are A1 and A2 independent events? Answer in two different ways. d. If you learn that the individual did not like vehicle #2, what now is the probability that he/she liked at least one of the other two vehicles QUESTION Consider randomly selecting a single individual and having that person test drive 3 different vehicles. Define events A1, A2, and A3 by A1 = likes vehicle #1 A2 = likes vehicle #2 A3 = likes vehicle #3. Suppose that P(A1) = 0.55, P(A2) = 0.65, P(A3) = 0.70, P(A1 ∪ A2) = 0.80, P(A2 ∩ A3) = 0.50, and P(A1 ∪ A2 ∪ A3) = 0.88. (a) What is the probability that the individual likes both vehicle #1 and vehicle #2? (b) Determine and interpret P (A2 |A3 ). (c) Are A2 and A3 independent events? Answer in two different ways. (d) If you learn that the individual did not like vehicle #1, what now is the probability that he/she liked at least one of the other two vehicles? SOLUTION The events are defined as follows: A₁ = an individual like vehicle #1 A₂ = an individual like vehicle #2 A₃ = an individual like vehicle #3 The information provided is: “P(A_1)=0.55\\P(A_2)=0.65\\P(A_3)=0.70\\P(A_1\bigcup A_2)=0.80\\P(A_2\bigcap A_3)=0.50\\P(A_1\bigcup A_2A_3)=0.88” Question (a) Compute the probability that the individual likes both vehicle #1 and vehicle #2 as follows: Solution “P(A_1\bigcap A_2)=P(A_1)+P(A_2)-P(A_1\bigcup A_2)\\=0.55+0.65-0.80\\Answer=0.40” Question (b) Determine and interpret P (A2 |A3 ). Solution “P(A_2|A_3)=\frac{P(A_2\bigcap A_3)}{P(A-3)}\\=\frac{0.50}{0.70}\\Answer=0.7143” Question (c) Are A2 and A3 independent events? Answer in two different ways. Solution If two events X and Y are independent then; “P(X\bigcap Y)=P(X)*P(Y)\\P(X|Y)=P(X)” The value of P (A₂ ∩ A₃) is 0.50. The product of the probabilities, P (A₂) and P (A₃) is: “P(A_2)*P(A_3)=0.65*0.70=0.455” Thus, P (A₂ ∩ A₃) ≠ P (A₂) × P (A₃) The value of P (A₂ | A₃) is 0.7143. The value of P (A₂) is 0.65. Thus, P (A₂ | A₃) ≠ P (A₂). The events A₂ and A₃ are not independent. Question (d) If you learn that the individual did not like vehicle #1, what now is the probability that he/she liked at least one of the other two vehicles? Solution “P(A_2 \bigcup A_3|A^C_1)=\frac{P((A_2\bigcup A_3)\bigcap A^C_1)}{P( A^C_1)}\\=\frac{0.88-0.55}{1-0.55}\\=0.7333” Thus, the probability that an individual likes at least one of A₂ and A₃ given they did not like A₁ is 0.7333. # What Will You Get? 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