For each of the situations described in this problem, first state the question i
ID: 3128796 • Letter: F
Question
For each of the situations described in this problem, first state the question in the form of the probability that is being requested, then do the computation. For example, if you were asked the probability of observing "heads" on each of 3 coin flips, you would say that the probability being requested is: p(Heads_1 and Heads_2 and Heads_3) = (.5)(.5)(.5) =.125. (a) If you rolled a pair of (fair) dice, what is the probability that both would come up "6"? (b) If you rolled a pair of dice, what is the probability that the first would be a 1 and the second would be a 5? (c) If you rolled a pair of dice, what is the probability that you would get a 1 and a 5? (d) If you rolled a pair of dice, what is the probability that they would sum to 6? (e) If you flipped a (fair) coin 5 times and the first 4 results were "tails," what is the probability that the fifth toss would also be "tails"? (f) If you drew 5 cards from a standard playing deck (i.e., 4 suits of 13 values each), what is the probability of drawing the 10, J, Q, K, and Ace of hearts in that order? (g) If you drew 5 cards from a deck, what is the probability that the first draw would be a 10 or J or Q or K or Ace of any suit? (h) If the first card you drew was a 10 of hearts, what is the probability that the second card would be a J, Q, K or Ace of hearts? (i) If the first 2 cards you drew were the 10 and J of hearts, what is the probability that the third card would be a Q, K or Ace of hearts? (j) If the first 3 cards were the 10, J and Q of hearts, what is the probability that the fourth card would be the K or Ace of hearts? (k) If the first 4 cards were the 10, J, Q, K of hearts, what is the probability that the four card would be the Ace of hearts? (I) What is the probability of drawing a royal flush in 5 cards (a royal flush is the 10, J, Q, K and Ace of the same suit?Explanation / Answer
a)
Note that the probability of a 6 is 1/6.
Hence, to get two sixes,
P(6, 6) = (1/6)*(1/6) = 1/36 [ANSWER]
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b)
Note that the probability of a 1 and 5 is 1/6.
So, for 1 and 5 in that order,
P(1 and 5) = (1/6)*(1/6) = 1/36 [ANSWER]
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c)
Note that the probability of a 1 and 5 is 1/6.
Hence, as we can reorder 1, 5, we multiply the product of probabilities by 2:
P(1 and 5) = 2*(1/6)*(1/6) = 1/18 [ANSWER]
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D)
There are 6*6 = 26 possibilities for a pair of dice.
There are 5 possibilities to get a sum of 6.
Hence,
P(sum of 6) = 5/36. [ANSWER]
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