Thread: conditional function probability

Hello,
I really need your help.

We choose randomly a number between from 0 to 3. This number indicates how many times we will throw a coin.
M = the chosen number
K = how many time we get heads

Determine the following probability functions for all values of the variables:
1) pM(m)
2) pK/M(k/2)
3) pM/K(m/2)
4) pK(k)
5) Determine the conditional probability function of N, if K is odd number


1) pM(0) = pM(1) = pM(2) = pM(3) = 1/4
2) Bayes’ theorem ??? [pK/M(k/2) =p(2/k) p(k) / p(2)]
3) Bayes’ theorem ???

Thanks in advance.

Originally Posted by astipd

Hello,
I really need your help.

We choose randomly a number between from 0 to 3. This number indicates how many times we will throw a coin.
M = the chosen number
K = how many time we get heads

Determine the following probability functions for all values of the variables:
1) pM(m)
2) pK/M(k/2)
3) pM/K(m/2)
4) pK(k)
5) Determine the conditional probability function of N, if K is odd number


1) pM(0) = pM(1) = pM(2) = pM(3) = 1/4
2) Bayes’ theorem ??? [pK/M(k/2) =p(2/k) p(k) / p(2)]
3) Bayes’ theorem ???

Thanks in advance.

For this I would write out the joint distribution and then just compute the required answers by summing the relevant terms of the joint distribution and renormalising if necessary by the probability of the conditioning event/s

So start by tabulating p(M,k) for all the possible (M,k)


CB

Thanks for your help.

I gave it a shot.

1) pM(0) = pM(1) = pM(2) = pM(3) = 1/4

2) M=2
HH
HT
TH
TT

pK/M(0/2) = 1/4
pK/M(1/2) = 1/2
pK/M(2/2) = 1/4
pK/M(3/2) = 0

3) K=2
HH
HHT
HTH
THH

pM/K(0/2) = 0
pM/K(1/2) = 0
pM/K(2/2) =1/4
pM/K(3/2) = 3/4


Any suggestions for the following parts?

There is a typing error in part 5), not N but M.