Tbh the question isnt complete .... does the probability remain constant for both cases ??
If not then there are 2 options:
Is the option putback after you find that its wrong or is it removed ??
Like: there are 10 cards, 6 are red .... you need to pick out a red one .... you take one card and its black ... do you keep it aside and pick from the rest or put it back, reshuffle and try again ???
The other mistake is .... wait .... Mnjul already said it
quote:
Originally posted by Mnjul
Because there's something important .Roy didn't say: Do you proceed to try to get such "something" even if you do get it in the first try?
Assuming that its like my cards eg and you dont put it back and once you get the correct answer you quit (Mnjuls point)
Probability of getting it in the 1st attempt: 6/10 (Region A -> P(A))
Probability of getting it in the 2nd attempt: 6/9 (Region B -> P(B))
So to get the correct answer you want a region that contains a point in A and be but not in both of them (not in the intersection
)
P(AvB) => P (A union B)
P(A<>B) => P (A intersection B)
(I didnt want any confusion as to the symbolisms .... cause the actual ones arent available 8-)
P(A<>B) = (6/10) * (5/9) --> (Condition where it happens both the times)
= 1/3
Answer:
P(AvB) - P(A<>B)
= [P(A) + P(B) - P(A<>B)] - P(A<>B)
= P(A) + P(B) - 2 * P(A<>B)
= (6/10) + (5/9) - (2/3)
= 22/45
= 0.4889
= 48.89 %
Note that this is the correct answer (afaik)
for the conditions: (Its been a year or so since i left probability
)
quote:
Originally posted by John Anderton in this post
Assuming that its like my cards eg and you dont put it back and once you get the correct answer you quit (Mnjuls point)
quote:
Originally posted by Mnjul
I'm too lazy to draw the tree diagram...anyone?
I was thinking about more of a regional diagram to support my answer but ill attatch them asap
It would be better if .Roy could tell us what he exactly wanted unless he wanted the answers in all aspects
/ 
