Marginalization of joint distribution

I am trying to understand how you marginalise a joint distribution.

In my case I have a fair coin, $P(C) = \frac12$ and a fair dice $P(D) = \frac16$. I am told I win a prize if I flip the coin and it lands on Tails and if the outcome of the dice $= 1$. I am told at least one of them is correct.

$$Q = (\text{Coin = Tails or Dice} = 1)$$

$$W = (\text{Coin = Tails and Dice} = 1)$$

So if I wanted to work out the probability $W =$ True $| Q =$ True I can use marginalisation to work this out given the joint distribution:

$$P(C), P(D), P(Q|C,D), P(W|C,D)$$

I am just not sure where to start any help would be really appreciated. I am pretty new to this.

Thanks in advance.

Topic bayesian-networks probability

Category Data Science

1
answered at

Guide:

To solve the problem, we have to assume that there is the outcome of the dice and the outcome of the coins are independent.

We let $C$ denotes the event that the coin lands on tail and $D$ be the event that the dice outcome is $1$.

You have been given that the coin lands on tails or the dice lands on $1$, and you are interested in finding out that the coin lands on tails and the dice lands on $1$.

You want to compute $P(W|Q)=\frac{P(W\cap Q)}{P(Q)}=\frac{P(W)}{P(Q)}=\frac{P(C)P(D)}{P(Q)}$.

To compute $P(Q)$ where $Q= C \cup D$. You can either use $$P(Q)= P(C)+P(D)-P(C\cap D)$$

or $$P(Q)=1-P(Q^c)=1-P(C^c \cap D^c)$$

Given all these formulas, hopefully you can solve for $P(W|Q)$.

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