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Probability question-flipping a coin-mathematical approach?

  1. Jun 7, 2012 #1
    1. What is the probability of tossing at least three 'heads' if a fair coin is tossed four times?

    2. Drawing a tree

    Hey guys,

    I was just wondering...is there a mathematical approach for me solving this question without drawing an event tree? Reason I ask is so that it saves time should a similar question appear on an exam

  2. jcsd
  3. Jun 7, 2012 #2
    You have 3 heads and 1 tail: H H H T.

    How many ways can you arrange these uniquely? It should be obvious that there are only four ways, but mathematically, you would take 4! (for the number of flips) and divide by 3! (the number of ways the heads can be switched among themselves that you can't distinguish) and divide by 1! (the same as before, but for the tails). 4!/3!/1! = 4. You might know this as [itex]\binom{4}{3}[/itex] also.

    Now, take 4, and multiply by the probability of each individual event. Each head occurs on its flip with .5 probability. .5^3 = .125. The same for tails. .5^3*.5 = .0625.

    The final result: .0625*4 = .25.
  4. Jun 7, 2012 #3
    ...but the OP's question was "at least three heads", so you also need to add on the probability of getting H H H H.
  5. Jun 7, 2012 #4
    So...I am more confused than ever, and understandably so. I thought the answer would be 5/16 but...I don't even know anymore
  6. Jun 7, 2012 #5
    5/16 is the correct answer. His answer of 25% is 4/16, but it doesn't include the possibility of H H H H, which is the extra 1/16.

    What part of his explanation confused you?
  7. Jun 7, 2012 #6
    Why "understandably so"? You may very well be "more confused than ever", but it's certainly not understandable that you are.

    You have been given explanation and the solution. :smile:
  8. Jun 8, 2012 #7
    Maybe looking at a slightly more general solution will help.
    I'll be talking about flipping k heads out of n flips. (So in this problem k is 3 and 4, and n is 4)

    [itex]\binom{k}{n}[/itex] is the number of ways that you can select which of the n flips will be heads. (since out of the n flips, k of them have to be choosen to land heads)
    So this is also the numerator of your probability, as it is the number of positive outcomes.
    The number of possible outcomes is [itex]2^{n}[/itex], because there are two possibilities (heads/tails) for each of the n flips.

    Thus the probability of getting k heads [or P(k,n)] would be [itex]\frac{\binom{k}{n}}{2^{n}}[/itex].

    In your problem, P(3,4) + P(4,4) = 1/4 + 1/16 = 5/16, which is the answer you got.
  9. Jun 8, 2012 #8

    Ray Vickson

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    Homework Helper

    How you proceed depends on what you already know. If you do not know about the binomial distribution, you need to work out the problem from first principles, perhaps by drawing a probability tree or something similar. If you know the binomial distribution formulas, you can use them directly, thus by-passing the probability tree or whatever.

    However, there is another issue. If (as in this case) we have an experiment in which the outcome has a binomial distribution taking possible values {0,1,2,...,N}, and if you are asked to compute a probability that at least a certain number occurs (or at most a certain number) you generally have no choice: you will have to perform a sum of several probabilities. The only exceptions are if you are asked "at least 1" or "at most N-1", because you can then look at the respective complementary events "exactly 0" or "exactly N" and get their probabilities instead (then subtract these from 1). In all other cases you need to do a sum (but the complementary probability may be easier because it may contain fewer terms).

  10. Jun 8, 2012 #9
    Thanks guys! I will have a look over this stuff to make sure I understand it.
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