Help with probability

Can any one help me to solve P(M=1/T=0)?

What's the answer???Is it 5/9 ???

Bayes theorem seems to do the trick.

@rahdirs, Thank you, for your help. I dont have the answer, but could you please explain to me how you came up with the answer?

By Bayes theorem, P(M=1/T=0) = P(T=0/M=1)*P(M=1)/P(T=0). We already know value of P(M=1) as 1/5.

So, we need to find value of P(T=0/M=1) & P(T=0);
P(T=0) = P(T=0/FS)*P(FS) (We know F & S are mutually exclusive,so P(FS)=P(F)*P(S))

therefore = P(T=0/FS)*P(FS) = P(T=0/F=0,S=0)*P(F=0,S=0) + P(T=0/F=1,S=0)*P(F=1,S=0) ----(I did not write other two cases of F=0,S=1 & F,S=1 as P(T=0,FS) is zero).

P(T=0) = 1*(2/5)*P(F=0) + (1/2)*(2/5)*P(F=1);

P(F=0) = P(F=0/M=0)P(M=0)+P(F=0/M=1)P(M=1) = (3/4)*(1/5) + (1/4)*(4/5) = 7/20;

P(F=1) = P(F=1/M=0)P(M=0)+P(F=1/M=1)P(M=1) = 1*(1/5) + 0*(4/5) = 1/5;

P(T=0) = 1*(2/5)*(7/20) + (1/2)*(2/5)*(1/5) = 9/50;

Similarly, P(T=0/M=1) = (3/4)*(2/5) + (1/2)*(2/5) = 10/20 = 1/2.

Thus, P(M=1/T=0) = P(T=0/M=1)*P(M=1)/P(T=0) = (1/5)*(1/2)/(9/50) = 5/9.

Do you see why i think the solution may be incorrect.

Dear rahdirs, Thank you for your extra ordinary replay. I agree with your explanation. While trying for myself I made mistake on expanding the expression
P(T=0/M=1) = P(T=0/F=0,S=0)*P(F=0/M=1)*P(S=0) +
P(T=0/F=0,S=1)*P(F=0/M=1)*P(S=1) +
P(T=0/F=1,S=0)*P(F=1/M=1)*P(S=0) +
P(T=0/F=1,S=1)*P(F=1/M=1)*P(S=1).
Thank you for clarifying this, Sir.

Can you calculate & post here the values of probability of P(F=0) & P(F=1) ?????
I have a query in that part.
I assumed that F & S are mutually exclusive and took P(FS) = P(F)*P(S).Are they exclusive ?????

You are right, since the dependency is shown by an arrow in directed graphs both F&S are are mutually exclusive.
Hence their probability will be the product of the P(F)*P(S). You already came up with expression for both P(F=0)& p(F=1) and I agree with the expression.
But I also see some problem since P(F=0)+P(F=1)!=1. I reasoned that has got to do with the dependency on 'M'. but let me know your opinion on that.
P(F=0) = P(F=0/M=0)P(M=0)+P(F=0/M=1)P(M=1) = (3/4)*(1/5) + (1/4)*(4/5) = 7/20;
P(F=1) = P(F=1/M=0)P(M=0)+P(F=1/M=1)P(M=1) = 1*(1/5) + 0*(4/5) = 1/5;

Yes, F & S seem to be mutually exclusive.

Yeah, P(F=0) + P(F=1) != 1.It may be dependent on M,but F still let's say an event,then sum of all its probable events must be 1 unless F has another outcome other than '0' & '1'.

Are 0 & 1,the only outcomes of F or are there other ???
Since P(F=0) + P(F=1) = 7/20 + 4/20 = 11/20,there must be some other outcome for x,or even more such that P(F=x) = 9/20.
Is that the case ????

Hi,
The outcomes of F are 0&1. if you need the detail. Here is the chart I collected the probability from based on the graph given on my first post.
M F S T
1 1 0 0
1 0 1 1
1 1 0 1
1 1 1 1
0 0 0 0

Bernoulli distribution is used to model the probabilities. As you can see there is only 0/1(binary) output. What do you think is the reason that P(F=1)+P(F=0)!=1?

I am not sure if the solution i posted is correct but i see no flaw in calculation of probability of F.

P(F=0) = P(F=0/M=0)*P(M=0) + P(F=0/M=1)*P(M=1) = (1/4)*(4/5) + (3/4)*(1/5) = 1/5 + 3/20 = 7/20.
P(F=1) = P(F=1/M=0)*P(M=0) + P(F=1/M=1)*P(M=1) = 0*(4/5) +1*(1/5) =1/5.

Are you sure that P(F=1/M=0) =0,if it is so & if F can only take values of '0' & '1',then i don't think there is an explanation for this error unless there is a flaw in above calculations & i don't think there is one.

Can you mark the thread as UNSOLVED so that other members of the forum can also comment.

The value of P(F=1/M=0) is zero. But why do you think it is important, since this is an event and we cant control the event. In addition,as it is shown on the network, even if F and M are dependent or independent, the probabilities of possible out comes need to be summed to one. I also don't think that small data-set(event table) could be the reason. I would like to know what you think. Thank you for the insight full discussion.