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Homework Help: Probability question i dont understand

  1. Sep 23, 2010 #1
    1. The problem statement, all variables and given/known data
    1. 2 card are chosen from a standard 52 card deck, what is the chances of getting the sum of them to be at least 4?
    aces = 1
    face cards = 10
    2. In a shuffled deck, what is the chances of ace of spades and ace of clubs to be adjacent?
    3.Someone flips 3 coins and rolls a fair die, what is the chances that the number of heads will equal the number showing on the die?

    2. Relevant equations

    3. The attempt at a solution

    1. Well i guess if the first card is not a ace or 2, then it desnt matter what the second card is, because then the sum of any combination will be (3+) + (1+) >= 4
    so then its 44/52 chance of not getting any of the aces or 2's.
    Since the ace of clubs can be anywhere, that leaves 51 cards left, so then there is a 1/51+1/51 = 2/51 chance of the ace of spades to be beside the ace of clubs.
    Last edited: Sep 23, 2010
  2. jcsd
  3. Sep 23, 2010 #2
    first off, let us think about the deck. There are 12 face cards, 4 aces, and then 4 sets of 2, 3, ..., 10

    So there are:
    sixteen 10s
    4 1s through 9s

    The best way to calculate the probability of having a sum greater than or equal to 4 is to calculate the probability of having a sum less than 4 and subtract it from 1.

    Our approach will be to count how many ways you can have a sum of 2 and of 3 and then divide those two added by the total number of sums possible. Finally, we will subtract this number from 1 to arrive to the answer.

    How many ways can your sum equal 2? First, I'd ask "does order matter?" Sort of. Choosing a then b is a different way to sum to (a + b) than choosing b then a. So we'll do permutations to be careful. Of all the cards, only 2 aces can add to 2. So we say the number of ways to add to 2 equals permutation(4, 2)

    How many ways can your sum equal 3? Again, order matters (because the different orders represent unique paths you could've taken to arrive to that sum, so it affects your probability of arriving to that sum), so we choose the permutations. The only way to sum to 3 is have a 1 and a 2 chosen. We could choose any 1 (4 possibilities) then any 2 (another 4 possibilities) but we could also choose any 2 (4 possibilities) then any 1 (another 4 possibilities)
    So the total ways to sum to 3 is 4*4*2 = 32

    So we now combine it all:
    -(32 + permuation(4,2))/permutation(52, 2)+1 = .9834
  4. Sep 23, 2010 #3
    OK i see now, but i still need help with 2 and 3..
  5. Sep 23, 2010 #4
    well i tried to use many words and describe the method so that you could try the approach out on the next two. the answer doesn't rely on some magical intuition of probabilities and cards like you were doing earlier. Instead, if you want to know the probability of doing X then either count the ways of doing X divided by the total ways to do it or count the ways not to do X divided by the total ways to do it and subtract it from 1.
  6. Sep 23, 2010 #5
    also for number 2 i am getting 2/52 and 1/1326, and i dont know which is right....

    I dont understand this question.
    Suppose we roll 3 fair die, what is the chances that 2 match and one is different.
    i put for this
    there are 216 outcomes
    chance of getting a number on die1 = 6/6
    chance of getting the same number as on die1 in die2 = 1/6
    chance of getting a different number as die1 in die3 = 5/6
    6/6 * 1/6 * 5/6 = 5/36
    Is this right?
    Last edited: Sep 23, 2010
  7. Sep 23, 2010 #6
    i figured out the previous questions

    but now i need help with these
    1. You are dealt 5 cards from a standard 52 card deck, what is the condition probability that the hand contains no pairs and no spades?
    2. The probability of snow is 20%, and the probability of a traffic accident is 10%. Suppose further that the conditional probability of an accident, given that it snows, is 40%. What is the conditional probability that it snows, given that there is an accident?
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