Stat1000W12_A4_sols

# Stat1000W12_A4_sols - STAT 1000 Assignment 4 – REVISED...

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Unformatted text preview: STAT 1000 Assignment 4 – REVISED DUE: March 14th (Wed. Eve. Section), March 15th (T/Th. Sections), March 16th (MWF. Sections) SHOW ALL YOUR WORK 1. [5] Suppose we have the following information about credit card ownership for Canadian adults: • 51% have a Mastercard (M) • 20% have an American Express card (A) • 70% have a Visa (V) • 29% have a Visa and a Mastercard • 14% have a Visa and an American Express card • 8% have a Mastercard and an American Express card • 6% have all three cards (a) If we randomly select one Canadian adult, what is the probability they have a Mastercard or a Visa? Solution: P(M or V) = P(M) + P(V) - P(M and V) P(M or V) = 0 . 51 + 0 . 70- . 29 = 0 . 92 (b) If we randomly select one Canadian adult, what is the probability that he or she has a Mastercard or an American Express, but not both? Solution: P(M or A) = P(M) + P(A) - P(M and A) P(M or A) = 0 . 51 + 0 . 20- . 08 = 0 . 63 Since we are looking for an individual who has either card but not both, 0 . 63- . 08 = 0 . 55 (c) What is the probability that a randomly selected Canadian adult has a Visa and no Mastercard? Solution: The easiest way to see this is by looking at the Venn diagram, "#\$% %&'(#)%* '+,('\$\$ &%\$-'()%(. /01/ /023 /04/ /015 /0/6 /037 /0/8 P(V) - P(V and M) . 70- . 29 = 0 . 41 (d) Are any two of the three events independent of one another? Explain? Solution: Two events are independent if and only if; P(A and B) = P(A)P(B) P(M and A) 6 = P(M)P(A) . 08 6 = (0 . 51)(0 . 20) . 08 6 = 0 . 10 Therefore, Mastercard and American Express are not independent. P(V and A) = P(V)P(A) . 14 = (0 . 70)(0 . 20) . 14 = 0 . 14 Therefore, Visa and American Express are independent. P(M and V) 6 = P(M)P(V) . 29 6 = (0 . 51)(0 . 70) . 29 6 = 0 . 357 Therefore, Mastercard and Visa are not independent. 2. [5] A baseball player compiles the following information: • He hits a homerun (H) in 34% of his games. • He gets a strikeout (S) in 40% of his games. • In 78% of his games, he hits a home run or his team wins (W). • In 10% of his games, he hits a home run and gets a strikeout. • In 26% of his games, he hits a home run and his team wins. • In 28% of his games, he gets a strikeout and his team wins. (a) In any given game, what is P(H or S)? Solution: P ( H or S ) = P ( H ) + P ( S )- P ( H and S ) = 0 . 34 + 0 . 40- . 10 = 0 . 64. (b) What is P(W)? Solution: We know P ( H or W ) = P ( H ) + P ( W )- P ( H and W ) ⇒ P ( W ) = P ( H or W )- P ( H ) + P ( H and W ) = 0 . 78- . 34 + 0 . 26 = 0 . 70. (c) What is P(H and W c )? Solution: The probability that the player gets a homerun is equal to the probability that he gets a homerun and the team wins, plus the probability that he gets a homerun and his team doesn’t win, i. e....
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## This note was uploaded on 04/01/2012 for the course STAT 1000 taught by Professor Xu during the Spring '12 term at Manitoba.

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Stat1000W12_A4_sols - STAT 1000 Assignment 4 – REVISED...

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