Probability of getting an odd number-conditional probability

[Fatima Hasan]

Homework Statement

Homework Equations

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The Attempt at a Solution

Probability of getting an odd number if the chosen dice was biased = P(O|B)
from the tree diagram , P(O|B) = 1/4
Could someone check my answer please ?

 

[member 587159]

You correctly identified on the tree diagram what $P(O|B)$ is. However, the probability is not $1/4$. The probability indicated on the tree diagram is the probability $P(O \cap B) = 1/4$.

Use the definition of conditional probability to find the correct answer.

 

[Fatima Hasan]

P(B) = 1/4 + 1/2 = 3/4
P(O∩B) = P(O|B)/P(B)
= (1/4 )/ (3/4) = 1/3
Right ?

 

[member 587159]

P(B) = 1/4 + 1/2 = 3/4
P(O∩B) = P(O|B)/P(B)
= (1/4 )/ (3/4) = 1/3
Right ?

Correct.

 

[Fatima Hasan]

You correctly identified on the tree diagram what $P(O|B)$ is. However, the probability is not $1/4$. The probability indicated on the tree diagram is the probability $P(O \cap B) = 1/4$.

Use the definition of conditional probability to find the correct answer.

Why it’s not conditional probability?

 

[member 587159]

Why it’s not conditional probability?

That's simply how probability trees should be read. The probability at the end is the probability of the intersection of all events that occurred to get to the end.

Another way to see that it isn't possible that this is conditional probability is to notice that $1/4 + 1/2 \neq 1$. Indeed, denote $\Omega$ for the sample space (universum). Then, $1 = P(\Omega|B) = P(E \cup O |B) = P(E|B) + P(O|B)$ ($E$ is the event that the result of the thrown dice is even, and $E\cap O = \emptyset$, so I can split it up) and according to your reasoning this should be equal to 3/4, which it clearly isn't.

EDIT: I also saw that in post 3, you made a mistake by applying the formula's:

It should be $P(O|B) = P(O\cap B)/P(B) = (1/4)/(3/4) = 1/3$

instead of $P(O\cap B) = P(O |B)/P(B)$ but I think this was a typo, because you obtained the correct answer.

 

[Ray Vickson]

Why it’s not conditional probability?

When you look at the branches leading out of the Biased die node, you see that one of them has twice the probability of the other. When we take a conditional probability (conditioned on a biased die) we look only at those two branches, and they keep their relative odds (2 to 1); thus, in the new sample space of "biased die only" one branch has probability 1/3 and the other has probability 2/3.

Another way to see this is to do a counting method. Say we do the whole experiment 8000 times. Of these 8000 times, (fair die, even number) and (fair die, odd number) each occur in 1000 of the experiments. The outcome (biased die, even number) occurs in (1/2)(8000) = 4000 trials, while (biased die, odd number) occurs in (1/4)(8000) = 2000 trials.

Altogether, we have 6000 trials with a biased die, of which 4000 of them give an even number and 2000 an odd number. When we restrict our attention to a biased die only we have 6000 experimental outcomes, and P(even|bias) = 4000/6000 = 2/3, P(odd|bias) = 2000/6000 = 1/3.