Probability, The AND

  1. adjacent

    adjacent 1,536
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    Probability, The "AND"

    Two coins are tossed.Find the probability of obtaining a hear OR a tail
    Ans:P(head)+P(tail)
    =1
    I understand this

    Again he tosses the coins.This time,what is the probability of obtaining a Head AND a tail?
    This time we multiply, P(head) X P(tail)
    =1/4

    But why do we multiply?What's the logic?
    P.S I understand the difference between AND and OR.
     
  2. jcsd
  3. tiny-tim

    tiny-tim 26,041
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    hi adjacent! :smile:
    the logic is that if the two events A and B are independent then we multiply the probabilities for AND:
    P(A and B) = P(A)*P(B)​

    if the two events A and B are mutually exclusive (non-everlapping), we add the probabilities for OR:
    P(A or B) = P(A) + P(B)​

    if the problem doesn't easily split into mutually exclusive events, then you may have to divide it up first

    for example, with two coins each tossed once, there are four mutually exclusive events:

    HH HT TH and TT​

    P(head AND tail) = P(HT OR TH) = P(HT) + P(TH) = 1/4 + 1/4 = 1/2

    P(head on the first die AND tail on the second die) =P(head on the first die)*P(tail on the second die) = 1/2*1/2 = 1/4

    (and P(head OR tail) = P(everything), so that's easy!

    P(head on the first die OR tail on the second die) =P(HT OR HH OR TT) = 1/4 + 1/4 + 1/4 = 3/4)
     
  4. Stephen Tashi

    Stephen Tashi 4,410
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    You are interpreting the phrase "a head or a tail" to refer to a sample space where an outcome is a single flip of a coin. You are interpreting the phrase "a head and a tail" to refer to a sample space where an outcome is 2 flips of a coin. So you are dealing with two different sample spaces.

    I suggest you try to phrase your question using an example where the "And" and "Or" of events uses events that are in the same sample space.
     
  5. adjacent

    adjacent 1,536
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    What do you mean by Sample space?
    I am using the same 2 coins over and over again.
     
  6. adjacent

    adjacent 1,536
    Gold Member

    And @tiny-tim
    Thank you,but..
    If the possible outcomes are HH HT TH TT,
    then the probability of A head AND a tail should be 1/2
    Then how is it 1/4? (1/2*1/2)
     
  7. tiny-tim

    tiny-tim 26,041
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    hi adjacent! :smile:
    it isn't!

    (i assume you mean the probability that, looking at the two coins, you see one head and one tail)

    are you confusing it with the different problem of the probability of a head on the first coin and a tail on the second coin (which are independent)? :wink:
     
  8. adjacent

    adjacent 1,536
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    Is specifically mentioning the head on the first coin and the second coin important?
    If so,What can be the answer for the question:What is the probability of obtaining a head and a tail?
    (No coins mentioned)
     
  9. tiny-tim

    tiny-tim 26,041
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    there are only two faces showing

    so "a head and a tail" must mean one head and one tail

    that's either HT or TH

    (if there were three coins, "obtaining a head and a tail" would mean at least one head and at least one tail, ie HHT HTH THH HTT THT or TTH)
    yes, that would be only HT instead of HT or TH
     
  10. adjacent

    adjacent 1,536
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    But according to AND rule,
    Probabilities of H and T is 1/4 since
    P(H) =1/2 .
    P(T) =1/2
    1/2*1/2=1/4
     
  11. tiny-tim

    tiny-tim 26,041
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    i don't understand what your H and T are :confused:

    (since there are two coins)
     
  12. adjacent

    adjacent 1,536
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    H is head and T is tail.So Probability of H and T means Either HT or TH.
    So that's 2/4=1/2 :confused:
     
  13. tiny-tim

    tiny-tim 26,041
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    yes, P of one head and one tail = P(HT OR TH)

    so that's P(HT) + P(TH) = 1/4 + 1/4 = 1/2, as you say

    but i don't understand what you meant by the following …
     
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  14. adjacent

    adjacent 1,536
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    ___________________
    I was just thinking of applying the AND rule without OR rule.
    Like this:
    Q-Find the probability of obtaining a Head AND a Tail.
    A coin has two sides of which one side is Head ,So:1/2 is the probability of obtaining a Head.
    A coin has two sides of which one side is Tail ,So:1/2 is the probability of obtaining a Tail.
    So The probability of obtaining a Head AND a Tail is 1/2*1/2
    =1/4

    I know I am wrong somewhere. :frown:
     
  15. tiny-tim

    tiny-tim 26,041
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    but if you talk about one coin, you must always say which coin you mean
    on which coin? :confused:

    (yes, i know it's the same on either coin, but you have to specify which coin or you can't go onto the next step)
     
  16. tiny-tim

    tiny-tim 26,041
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    let's try this …

    there are four events each of which has probability 1/2 …

    P(1=H) = P(1=T) = P(2=H) = P(2=T) = 1/2 …

    which of those four events were you trying to combine to find "the probability of obtaining a Head AND a Tail"? :smile:
     
  17. adjacent

    adjacent 1,536
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    Let's go to the first Question in the OP.
    "He tosses the coins.What is the probability of obtaining a Head AND a tail?"
    So that means no coin is specified.
     
  18. D H

    Staff: Mentor

    The question in the OP was ambiguous. Does it mean
    • What is the probability of obtaining heads on the first flip and tails on the second or
    • What is the probability of obtaining one heads and one tails on the two flips?
    With the first interpretation, the answer is 1/4. With the second, it's 1/2.
     
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  19. Stephen Tashi

    Stephen Tashi 4,410
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    I mean the space of outcomes for an experiment. If you don't understand that, I think you should study how the phrase is used in probability theory. http://en.wikipedia.org/wiki/Sample_space
     
  20. HallsofIvy

    HallsofIvy 40,804
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    Your use of the "and rule" is incorrect. What is true is that if P and Q are independent events, then the probability that P and Q happen in that order is P(P)P(Q). The probability that Q and P happen in that order is P(Q)P(P). The probability that P and Q happen in either order is P(P)P(Q)+ P(Q)P(P)= 2P(P)P(Q)
     
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  21. adjacent

    adjacent 1,536
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    Thank you so much everyone,I have got the answer now.I was just starting with probability.
     
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