By Ross S., Weatherwax J.L.
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Additional resources for A solution manual for A first course in probability
Then the location of the first pair in ways, and finally the 4 1 , the location of the second pair in 6 2 3 2 to select the third number. When we multiply these we get 6 2 5 2 3 2 4 1 = 1800 . 2315 . 65 Part (d): We can pick the number for the digit that is repeated three times in six ways, another digit in five ways and the final digit in four ways. The number of ways we can 5 place the three dice with the same numeric value is given by ways. So the number of 3 permutations of these three numbers is given by 5 3 6·5·4· This gives a probability of 1200 65 = 1200 .
4 1 . 18 Part (b): The event that Rebecca and Elise are chosen and then do not play each other will occur with a probability of P (R)P (E)P (Pairedc |R, E) = 1 4 1 1− · 2 9 4 = 1 . 6 Part (c): For this part we can have either (and these events are mutually exclusive) Rebecca picked to represent her school or Elise picked to represent her school but not both and not 7 1 is the number of ways to choose the team A with Rebecca as neither. Since 3 1 8 are the number of ways to choose team B without having Elise as a a member and 4 member, their product is the number of ways of choosing the first option above.
323 Problem 34 (the probability of a Yarborough) We must not have a ten, a jack, a queen, a king, or an ace (a total of 5 face cars) in our hand of thirteen cards. The number of ways to select a hand that does not have any of these cards is equivalent to selecting thirteen cards from among a set that does not contain any of the cards mentioned above. 000547 , a relatively small probability. 8363 . Where in the numerator we have enumerated all possible selections of three people such that at least one psychologist is chosen.