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Coin flip and dice roll question (check my answers?)

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data

    Hi,
    I just wanted to check my answers for this question:

    Q: Suppose that we flip a coin six times and roll a 6-sided die once. Suppose also that all outcomes of this experiment (consisting of an ordered sequence of results for the flips (heads or tails) and the number showing on the die after the roll (an integer from 1 to 6)) are equally likely.
    Find the following probabilities:
    1) The probability that all six flips are heads and the die shows the number 6
    2) The probability that the first R flips are heads, where R is whatever number is showing on the die.
    3) The probability that the number of flips that are heads times the number showing on the die is 18.
    4) The probability that the number of heads in a the six flips is same as the number showing on the die.

    2. Relevant equations



    3. The attempt at a solution

    A:
    1) Let A be that event that all six flips are heads.
    Let B be the event that the dice roll is 6.
    P(A) = #A/#S, #B = 1, #S = 2^6, so P(A) = 1/64
    P(B) = #B/#S, #b = 1, #S = 6, so P(B) = 1/6
    Since A and B are independent, P(A)∩P(B) = P(A)P(B) = 1/384

    2) We want the roll on the die and the number of heads in 6 flips to match.
    The chance to get any roll is always 1/6.
    The chance to get first as head is 2^5 / 64.. first 2 as heads is 2^4/64.. etc. Each is multiplied by 1/6 (because 'and').
    Let {P(A),...,P(F)} be these probabilities.
    Any of them happening will suffice.
    So let E be the event in the question.
    P(E) = P(A) + ... + P(F) (because 'or')
    = (1/6)((32+16+8+4+2+1)/64) = 63/384

    3) Let E be the event.
    This can only happen if we roll 3 and get 6 heads or roll 6 and get 3 heads.
    Let A and B be those events, so P(E) = P(A) + P(B)
    P(A) = roll 3 and 6 heads = (1/6)(1/64) = 1/384
    P(B) = roll 6 and 3 heads = (1/6)(2^3/64) = 20/384
    So P(E) = 21/384

    4) We want number of heads and roll to be the same.
    So, 1 head and roll 1 or 2 head and roll 2 or ... or 6 head and roll 6.
    Rolling any number is 1/6. Permutation for 1 head any order is 6 choose 1.. 2 head is 6 choose 2 etc. All possible is 2^6 = 64.
    Let E be the event, then P(E) = (1/384)(6 + 15 + 20 + 15 + 6 + 1) = 63/384

    Thanks
     
  2. jcsd
  3. Oct 6, 2011 #2
    Just one thing - for question 3 you've got 2^3 = 20. Should be 8.
     
  4. Oct 6, 2011 #3

    vela

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    I think zeion just typed his work in incorrectly. Something else should be there instead of 2^3, but his final answer is correct.
     
  5. Oct 6, 2011 #4
    you're right vela - 6 choose 3 gives 20.
     
  6. Oct 6, 2011 #5
    Yes I meant 6 choose 3 thanks guys.
     
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