STAT 400 Lecture AL1 Answers for 1.1 Spring 2015 Dalpiaz 1. Suppose a 6-sided die is rolled. The sample space, S , is { 1, 2, 3, 4, 5, 6 }. Consider the following events: A = { the outcome is even }, B = { the outcome is greater than 3 }, a) List outcomes in A, B, A', A ∩ B, A ∪ B. A = { the outcome is even } = { 2, 4, 6 }, B = { the outcome is greater than 3 } = { 4, 5, 6 }, A' = { 1, 3, 5 }, A ∩ B = { 4, 6 }, A ∪ B = { 2, 4, 5, 6 }. b) Find the probabilities P( A ), P( B ), P( A' ), P( A ∩ B ), P( A ∪ B ) if the die is balanced (fair). P( A ) = 3/6 , P( B ) = 3/6 , P( A' ) = 3/6 , P( A ∩ B ) = 2/6 , P( A ∪ B ) = 4/6 .c) Suppose the die is loaded so that the probability of an outcome is proportional to the outcome, i.e. P( 1 ) = p, P( 2 ) = 2 p, P( 3 ) = 3 p, P( 4 ) = 4 p, P( 5 ) = 5 p, P( 6 ) = 6 p. i) Find the value of p that would make this a valid probability model. P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1. p + 2 p + 3 p + 4 p + 5 p + 6 p = 21 p = 1. ⇒ p = 1/21 . ii) Find the probabilities P( A ), P( B ), P( A' ), P( A ∩ B ), P( A ∪ B ). P( A ) = P(2) + P(4) + P(6) = 2/21 + 4/21 + 6/21 = 12/21 . P( B ) = P(4) + P(5) + P(6) = 4/21 + 5/21 + 6/21 = 15/21 . P( A' ) = 1 – P( A ) = 1 – 12/21 = 9/21 . OR P( A' ) = P(1) + P(3) + P(5) = 1/21 + 3/21 + 5/21 = 9/21 . P( A ∩ B ) = P(4) + P(6) = 4/21 + 6/21 = 10/21 . P( A ∪ B ) = P(2) + P(4) + P(5) + P(6) = 2/21 + 4/21 + 5/21 + 6/21 = 17/21 . OR P( A ∪ B ) = P( A ) + P( B ) – P( A ∩ B ) = 12/21 + 15/21 – 10/21 = 17/21 .2. Consider a “thick” coin with three possible outcomes of a toss ( Heads, Tails, and Edge ) for which Heads and Tails are equally likely, but Heads is five times as likely than Edge. What is the probability of Heads ? P ( Heads ) = P ( Tails ) = p for some p. P ( Edge ) = 51 p. P ( Heads ) + P ( Tails ) + P ( Edge ) = 1. p + p + 51 p = 1. 511 p = 1. P ( Heads ) = p = 115. 3. The probability that a randomly selected student at Anytown College owns a bicycle is 0.55, the probability that a student owns a car is 0.30, and the probability that a student owns both is 0.10. P( B ) = 0.55, P( C ) = 0.30, P( B ∩ C ) = 0.10. a) What is the probability that a student selected at random does not own a bicycle? P( B' ) = 1 – P( B ) = 1 – 0.55 = 0.45. C C ' B 0.10 0.45 0.55 B' 0.20 0.25 0.45 0.30 0.70 1.00b) What is the probability that a student selected at random owns either a car or a bicycle, or both? P( B ∪ C ) = P( B ) + P ( C ) – P( B ∩ C ) = 0.55 + 0.30 – 0.10 = 0.75. OR P( B ∪ C ) = P( B ∩ C ) + P( B' ∩ C ) + P( B ∩ C' ) = 0.10 + 0.20 + 0.45 = 0.75. OR P( B ∪ C ) = 1 – P( B' ∩ C' ) = 1 – 0.25 = 0.75. c) What is the probability that a student selected at random has neither a car nor a bicycle? P( B' ∩ C' ) = 0.25. 4. During the first week of the semester, 80% of customers at a local convenience store bought either beer or potato chips (or both). 60% bought potato chips. 30% of the customers bought both beer and potato chips. What proportion of customers bought beer? P( B ∪ PC ) = 0.80, P ( PC ) = 0.60, P( B ∩ PC ) = 0.30. P( B ∪ PC ) = P( B ) + P ( PC ) – P( B ∩ PC ). 0.80 = P( B ) + 0.60 – 0.30. ⇒ P( B ) = 0.50.5. Suppose P ( A ) = 0.22, P ( B ) = 0.25, P ( C ) = 0.28, P ( A ∩ B ) = 0.11, P ( A ∩ C ) = 0.05, P ( B ∩ C ) = 0.07, P ( A ∩ B ∩ C ) = 0.01. Find the following: a) P ( A ∪ B ) b) P ( A' ∩ B' ) c) P ( A ∪ B ∪ C ) d) P ( A' ∩ B' ∩ C' ) e) P ( A' ∩ B' ∩ C ) f) P ( ( A' ∩ B' ) ∪ C ) g) P ( ( A ∪ B ) ∩ C ) h) P ( ( B ∩ C' ) ∪ A' ) a) P ( A ∪ B ) = 0.36. b) P ( A' ∩ B' ) = 0.64. c) P ( A ∪ B ∪ C ) = 0.53. d) P ( A' ∩ B' ∩ C' ) = 0.47. e) P ( A' ∩ B' ∩ C ) = 0.17. f) P ( ( A' ∩ B' ) ∪ C ) = 0.75. g) P ( ( A ∪ B ) ∩ C ) = 0.11. h) P ( ( B ∩ C' ) ∪ A' ) = 0.88.6. Let a > 2. Suppose S = { 0, 1, 2, 3, … } and P( 0 ) = c, P( k ) = ka 1, k = 1, 2, 3, … . a) Find the value of c ( c will depend on a ) that makes this is a valid probability distribution. Must have ()∑xxp all = 1. ⇒ c + ∑∞=1 1 kka = 1. bbkk−=∞∑=11 0 , | b | < 1. 11111111 1 01−=−−=−∞=∞∑∑==aaaakkkk. OR 1111111 11 01−=−=∞=∞⋅⋅∑∑==aaaaaakkkk. c + 11−a = 1. c = 1212111−−=−−=−−aaaaa. b) Find the probability of an odd outcome. P ( odd ) = p ( 1 ) …
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