Probability Puzzle

PGM
PGM Registered Posts: 1,954 Beyond epic contributor ๐Ÿง™โ€โ™‚๏ธ
I thought I was fairly good at maths and probabilities until I tried this GCSE one, and got stuck....

Bit embarrasing really.

A coin and a dice are thrown at the same time.

What is the probability that it will be either heads and/or more than 4.

Comments

  • A-Vic
    A-Vic Registered Posts: 6,970 Beyond epic contributor ๐Ÿง™โ€โ™‚๏ธ
  • Claire321
    Claire321 Registered Posts: 209 Dedicated contributor ๐Ÿฆ‰
    I would say probability is 3/12 or 25%
  • crispy
    crispy Registered Posts: 466 Dedicated contributor ๐Ÿฆ‰
    I reckon:

    50 % * 33.333 % = 16.7 %

    or

    0.5 * (2/6) = 0.166667 = 16.7 %

    Or I could be completely wrong.
  • CJC
    CJC Registered Posts: 1,657 Beyond epic contributor ๐Ÿง™โ€โ™‚๏ธ
    Probability was never my strong point and I've no idea if this is right. but I'd work it like this

    The probability of a head is 1/2 and the probability of being more than 4 is 1/3.

    p(A AND B) = pA x pB so the probability of being a head AND over 4 is 1/2 x 1/3 = 1/6 (or 16.7%).

    p(A OR B) = (1 - p(~A AND ~B) = 1-p(~A) x p(~B), so the probability of being a head OR over 4 is 1 - (1/2 x 2/3) = 2/3 (or 66.67%)
  • Monsoon
    Monsoon Registered Posts: 4,071 Beyond epic contributor ๐Ÿง™โ€โ™‚๏ธ
    oo i rememer some of how to do this

    . 5 or more 2/6
    . /
    H 1/2 \
    /
    \
    T 1/2

    multiple 1/2 by 2/6 (i think) and the answer is 0.166 recurring

    Note I've taken "more than 4" to be 5 or 6, not 4, 5 or 6.

    o wait, it was and/or..... i dont have time for this, i have a maths GCSE already :lol:

    edit it wont allow the formatting but i remember doing branch type things....
  • Marga
    Marga Registered Posts: 981 Epic contributor ๐Ÿ˜
    there are thrown at the same time so they are independent

    CJC is correct with the result :)

    i love probability and stats!
  • PGM
    PGM Registered Posts: 1,954 Beyond epic contributor ๐Ÿง™โ€โ™‚๏ธ
    CJC wrote: ยป

    p(A OR B) = (1 - p(~A AND ~B) = 1-p(~A) x p(~B), so the probability of being a head OR over 4 is 1 - (1/2 x 2/3) = 2/3 (or 66.67%)

    Thats the only right answer on here!

    Official answer to the GCSE question is 2/3 (or in other words .66667)

    It confused me how you have to do 1 minus then the probabilities of it not being A or B. Or else in effect you end up double counting some of the outcomes. Brain baffling!

    Oh and well done you annoyingly smart person!
  • Toffeemadblue
    Toffeemadblue Registered Posts: 102 Dedicated contributor ๐Ÿฆ‰
    I couldn't remember any probability theory so I tried it using inspection as there only limited outcomes.
    H1,H2,H3,H4,H5,H6
    and
    T1,T2,T3,T4,T5,T6

    So twelve in all, of which eight satisfy the condition
    8/12 = two thirds
  • PGM
    PGM Registered Posts: 1,954 Beyond epic contributor ๐Ÿง™โ€โ™‚๏ธ
    I couldn't remember any probability theory so I tried it using inspection as there only limited outcomes.
    H1,H2,H3,H4,H5,H6
    and
    T1,T2,T3,T4,T5,T6

    So twelve in all, of which eight satisfy the condition
    8/12 = two thirds

    Yes thats right also, thats the way I did it. I wanted to know the way to calculate it also.
  • CJC
    CJC Registered Posts: 1,657 Beyond epic contributor ๐Ÿง™โ€โ™‚๏ธ
    I started out using inspection, got the answer that way and then worked backwards to figure out the maths (then dressed it up in a bit of notation to make it look clever!).
  • Marga
    Marga Registered Posts: 981 Epic contributor ๐Ÿ˜
    CJC wrote: ยป
    I started out using inspection, got the answer that way and then worked backwards to figure out the maths (then dressed it up in a bit of notation to make it look clever!).

    basically probability follows logic or if you want the operational maths signs

    Logically:

    P v Q = 7[(7P) ^ (7Q)]

    where v = OR ^ = AND 7P means OPPOSITE of P and the rules apply as per
    7(7P) = (77)P = P (two negatives = one positive)
    and 7v = ^ and 7^=v

    thats why the probability of an OR is the oppossite of the probability of none of the experiments happening

    With operational maths signs:

    a - b = - [ (-a) + (-b)] where - = OR and += AND
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