{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

quiz1sol_Real - PROBABILITY AND DECISION MAKING 45-730 QUIZ...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
PROBABILITY AND DECISION MAKING 45-730 QUIZ 1 (OFFLINE) SOLUTIONS Problem 1 ( Sample Spaces ) . Four friends toss an unbiased coin each to determine a winner. The per- son whose coin shows up different from all the other three wins. The outcome of any toss is independent of the others. (1.1) What is the sample space of this random experiment? Solution. There are 2 4 = 16 possible outcomes in tossing four coins. / (1.2) What is the probability that the four coins show up all heads or all tails? Solution. The probability that the four coins show up all heads is 1 16 , and there is the same probability that they show up all tails. Thus the probability that the four coins show up all heads or all tails is 1 16 + 1 16 = 1 8 . / (1.3) What is the probability that a coin shows up different from all the other three? Solution. The only outcomes where we have one coin different from all other three are: HTTT, THTT, TTHT, TTTH,THHH, HTHH, HHTH, HHHT. So the probability is 8 16 = 1 2 . / Problem 2 ( Sample Spaces and Set Theory ) . Two games are supposed to take place between teams A and B. The probability that team A wins the first game is 0.3, and the probability that it wins the second is 0.5. Suppose the probability that team B wins BOTH games is 0.5. (2.1) What is the appropriate sample space to analyze this problem? Solution. An appropriate sample space would be to consider the outcomes of both games. So we have AA, AB, BA, BB , where AB represents the event that team A wins the first game and B the second. For other outcomes is similar. The events AA, AB, BA, BB are mutually exclusive and collectively exhaustive. / (2.2) What is the probability of team A winning both games? Solution. From the data we have P ( BB ) = 0 . 5, P ( AA BA ) = 0 . 5 and P ( AA AB ) = 0 . 3. Note that P ( BB )+ P ( AA BA ) = 1 . 0, and this implies that we must have P ( AB ) = 0 . 0. Since the considered events are mutually exclusive. This implies that P ( AA ) = 0 . 3. / (2.3) Are the events that Team A wins the first game, and Team A wins the second game independent? Solution. Let A # be the event that Team A wins the first game and # A the event that Team A wins the second game. A # and # A are not independent since the condition P ( AA ) = P ( A # # A ) = P ( A #) · P (# A ) does not hold. / (2.4) What is the probability of B winning the first game? Solution. From 2.2 we have that P ( AA ) = 0 . 3. So we get that P ( BA ) = 0 . 5 - 0 . 3 = 0 . 2. The probability of B winning the first game is P ( BA ) + P ( BB ) = 0 . 2 + 0 . 5 = 0 . 7. / (2.5) What is the probability of B winning the second game? Solution. We have that P ( AA ) = 0 . 3, P ( BA ) = 0 . 2, P ( AB ) = 0 . 0 and P ( BB ) = 0 . 5. So the probability that B is winning the second game is 0 . 5. / 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2 QUIZ 1 (OFFLINE) SOLUTIONS Problem 3 ( Counting I ) . We draw at random two cards out of a standard deck of 52 cards without replacement. (3.1) What is the probability that the first card is red (the second can be red or black)? Solution. Since the order matters, the size of the sample space is the number of possibilities for the first card (52) times the number of possibilities for the second (51, since the first card is not replaced). Of this the number where the first card is red (and the second card is any of the remaining) is 26 times 51. The required probability is
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}