Calculating the Probability of a Specific Outcome from Throwing 3 Coins

[nhrock3]

we throw 3 coins

y represents the throw of three coins
x represents the throw of the first two
we count the number of hetz gotten from the thrown coins.

the solution says for the slot that:
P(x=1,y=2)=0.25
but i can't understand how
?
the logical solution says:
for x to be 1 we have 01 10
so for y to be 2 011 101
so with respect to y we have two possibilities from 8

but the formal mathematical says

P(x=1,y=2)=P(x=1)*P(y=2/x=1)
for P(x=1) we need to have 01 10 which is 1/2 or 0.5
for P(y=2/x=1) its 2/8 or 0.25

P(x=1,y=2)=P(x=1)*P(y=2/x=1)=0.5*0.25=0.125
and not 0.25

 

[Mulder]

but the formal mathematical says

P(x=1,y=2)=P(x=1)*P(y=2/x=1)
for P(x=1) we need to have 01 10 which is 1/2 or 0.5
for P(y=2/x=1) its 2/8 or 0.25

looks like the only mistake is in this last line. p(y=2|x=1) would be 1/2

the "logical" solution is fine, and the best way to do the problem.

 

[nhrock3]

why p(y=2|x=1) 1/2 ?
3 digit gives 8 possibilities
our possibilities are 011 101
so its 1/4

 

[vela]

You're mixing up p(y=2 and x=1) and p(y=2|x=1). The probability p(y=2 and x=1) is 2/8, but p(y=2|x=1) = p(y=2 and x=1)/p(x=1) = (2/8)/(1/2) = 1/2.

Intuitively, when you're given x=1, the possible outcomes are restricted to 010, 011, 100, and 101, so you divide by 4 instead of 8.

 

[nhrock3]

thanks i understand now