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Homework Help: If 4 coins are tossed, find the probabilities

  1. Sep 28, 2009 #1
    If 4 coins are tossed, find the following probability:

    2 heads.

    more than 3 tails.

    For 2 heads I got 1/16.
    More than 3 heads I don't know how to start that problem.
  2. jcsd
  3. Sep 28, 2009 #2
    Hi rowdy3,

    First I need to ask do you mean "more that 3 heads" or "more than 3 tails" as you wrote both in you question :D, ill assume for now that you actually meant more than 3 heads.

    So you two probabilities would be:

    P1 : exactly 2 heads
    P2 : more than 3 heads (this could also be written "exactly 4 heads"!)

    now first how did you work out exactly 2 heads was 1/16. Show us how you got to that answer , is that in fact correct? then we can help you further :D

    Are you using a probability tree, as if you are trying to work out the probability by thinking about it, you may get the incorrect answer, a probability tree will help you visualise the question much easier.
  4. Sep 28, 2009 #3
    I did it wrong. I should do 4c2/ 2^4 which will come out to 6/16 for 2 heads.
    More than 3 tails I don't know how.
  5. Sep 28, 2009 #4


    Staff: Mentor

    With 4 coins, "more than 3 tails" means "3 tails" or "4 tails." Can you find the probability of each? Since these are independent events, their probabilities add.
  6. Sep 28, 2009 #5
    more than 3 heads.

    odds of 1 coin landing on tails 1/2
    odds of 2 coins landing on tails 1/2 * 1/2 = 1/4
    odds of 3 coins tails 1/2 * 1/2 * 1/2 = 1/8
    odds of 4 coins tails 1/2 * 1/2 * 1/2* 1/2 = 1/16
  7. Sep 28, 2009 #6


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    Homework Helper

    not quite - i would re-write what you have as
    T - odds of 1st coin landing on tails 1/2
    TT - odds of 1st & 2nd coins landing on tails 1/2 * 1/2 = 1/4
    TTT - odds of 1st & 2nd & 3rd coins tails 1/2 * 1/2 * 1/2 = 1/8
    and so on...

    now consider this case when the coin is tossed 4 times, with only the last toss being a tail, ie.
    the probabilty of this event (1/2)^4 = 1/16

    Any individual ordered outcome will have a discrete probabilty of 1/16, as there are 16 different ordered outcomes...
    so there are 4 different ways to get a single tail
    so the total probabilty of a single head will be 4*(1/16) = 1/4

    so for the case where you want to find 2 heads, you must sum the probability of all the cases where you have two heads... eg.
    any good ideas on how to count the cases?
    Last edited: Sep 28, 2009
  8. Sep 28, 2009 #7
    I'm sorry, it's more than 3 tails. Would it still be 1/16?
  9. Sep 28, 2009 #8


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    Homework Helper

    no, any single outcome has probabilty of 1/16,

    you need to count all the possible outcomes with 3 or 4 tails...

    (note that it may be easier to consider the cases of 1 or less heads, which is equivalent)
  10. Sep 29, 2009 #9
    Hi guys, I thought Id comment again, surly this is wrong, if the probability was:

    "Find the probability of 3 or more tails" (thanks rowdy for clarify which ;D)

    then yes it would be the probability of exactly 3 heads (all permutations) and exactly 4 heads (only one permutation), but its not the probability as rowdy wrote it is:

    "Find the probability of more than 3 tails"

    which is exactly 4 tails, NOT 3 tails (all permutations), in which case rowdy you would be correct with the probability being 1/16, am I crazy or something :D But rowdy awesome on the first one, well done for spotting the mistake.
  11. Nov 26, 2009 #10
    a) 6/16
    b) 1/16
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