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Thread: Coin and Die probability

  1. #1
    ble
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    Coin and Die probability

    Question: A fair coin is tossed once and a fair die is rolled once. Calculate the probability of observing that the coin is heads or dice is six, P(R=True). NOTE: R in this case is for the event that coin is heads or dice is six.

    My Solution:


    Let A be the event that the coin toss results in a head.
    Let B be the event that the roll of the die results in 6.

    The probability of at least one of A or B' is:

    P(A∪B)=P(A)+P(B)−P(A∩B)

    And since A and B are independent,

    P(A and B)=P(A)⋅P(B)

    P(A or B)= 1/2 + 1/6 − 1/2 x 1/6
    =2/3−1/12
    =7/12


    The solution requires formal calculation using mathematical notation and the derivation of every step in the calculation. Since its a 10 mark question, I believe that my solution is not enough to get the full marks in an exam. Is there another approach that I can take to get a better answer? Should I use the Bayes theorem formula for this question?

    Thank you in advance.
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  2. #2
    MHF Contributor

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    Re: Coin and Die probability

    Quote Originally Posted by ble View Post
    Question: A fair coin is tossed once and a fair die is rolled once. Calculate the probability of observing that the coin is heads or dice is six,
    Let A be the event that the coin toss results in a head.
    Let B be the event that the roll of the die results in 6.
    P(A∪B)=P(A)+P(B)−P(A∩B)
    And since A and B are independent,
    P(A and B)=P(A)⋅P(B)
    P(A or B)= 1/2 + 1/6 − 1/2 x 1/6
    =2/3−1/12
    =7/12
    The solution requires formal calculation using mathematical notation and the derivation of every step in the calculation. Since its a 10 mark question, I believe that my solution is not enough to get the full marks in an exam. Is there another approach that I can take to get a better answer? Should I use the Bayes theorem formula for this question?
    If events $A~\&~B$ are independent then $\mathscr{P}(A\cup B)=\mathscr{P}(A)+\mathscr{P}(B)-\mathscr{P}(A)\cdot\mathscr{P}(B)$
    So your notation is fine in my classes.
    Thanks from ble
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  3. #3
    MHF Contributor

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    Re: Coin and Die probability

    Darn! You started out using the correct singular "die" but then changed to the incorrect "dice"!

    The coin is heads:
    H1
    H2
    H3
    H4
    H5
    H6

    The die is "6":
    H6
    T6

    There are a total of 6+ 2 except that "H6" has been counted twice. There are 6+ 2- 1= 7 successful outcomes out of a total of 2*6= 12 possible outcomes. The probability is 7/12. That is, of course, exactly the same as your "P(A)+ P(B)- P(AB)= 6/12+ 2/12- 1/12.
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