Probability Puzzle
PGM
Registered Posts: 1,954 Beyond epic contributor ๐งโโ๏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.
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.
0
Comments


I would say probability is 3/12 or 25%0

I reckon:
50 % * 33.333 % = 16.7 %
or
0.5 * (2/6) = 0.166667 = 16.7 %
Or I could be completely wrong.0 
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 = 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 = (1  p(~A AND ~B) = 1p(~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%)0 
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
edit it wont allow the formatting but i remember doing branch type things....0 
there are thrown at the same time so they are independent
CJC is correct with the result
i love probability and stats!0 
p(A OR = (1  p(~A AND ~B) = 1p(~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!0 
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 thirds0 
Toffeemadblue wrote: ยปI 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.0 
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!).0

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 += AND0
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