Joint Probability Vs Conditional Probability

I was studying about Joint probability and Conditional probability. From the reference below link is suggesting that we can easily find out the joint probability using conditional probability formula. It's just a cross-checking if the conditional probability formula is valid for joint probability as well.

My question is: A die is tossed, suppose A is the event that a prime number occurs, B is the event than an even number occurs. Find probability that prime number occurs when even turns up.

This is definitely a conditional probability question...

We need to find P(A|B)=?

P(A)= 3/6

P(B)= 3/6

Formula: P(A|B) = P(A and B) / P(B)

I know if I write the set A, B which are A ={2,3,5} and B={2,4,6} then P(A and B) is 1/6 as there is one common "2" from both sets. But if I use joint probability formula which is:

P(A and B) = P(A).P(B) so, the answer is wrong... (9/36)

However, using conditional probability formula I get a different answer for P(A and B)

P(A and B) = P(A|B).P(B) = (1/3).(3/6) = 1/6

Please let me know why I am getting different answers for P(A and B) when using joint probability and conditional probability formula

• $P(A\cap B)=P(A)P(B)$ is not joint-probability formula, it's independence formula. Of course, $A$ and $B$ are not independent here – user160738 Mar 6 '18 at 9:23
• I found the formula from below source: investopedia.com/terms/j/jointprobability.asp – muhammad tayyab Mar 6 '18 at 9:26
• This is what source said: You can also use a formula to calculate the joint probability – P(6 ∩ red) = P(6) x P(red) = 4/52 x 26/52 = 1/26. – muhammad tayyab Mar 6 '18 at 9:27

Your computation of conditional probability sounds ok. P(A and B) = 1/6 for the reason you state.

So the mistake is in the sentence: 'P(A and B) = P(A) and P(B) so, the answer is wrong... (9/36)'

There are actually two mistakes. First 'P(A) and P(B)' doesn't mean anything, from the remainder of the sentence we can infer that you mean 'P(A and B) = P(A) times P(B)'.

However: this does only hold when the events are independent. For instance, when you throw two dice (one red, one green) and you want the probability that the red die gives a prime number and the green one gives an even number.

Here however, with one die, there is no independence between A and B and you can't use the formula for independent events

The second is okay.

Your main mistake is "P(A and B)=P(A) and P(B)" where you probably mean something like: $$P(A\cap B)=P(A)\times P(B)\tag1$$ which in this case is simply not true.

Formula $(1)$ is only valid if $A$ and $B$ are independent.

Note that the events $A$ and $B$ both occur if and only if the die shows a $2$, leading to $P(A\cap B)=\frac16$.

This corresponds with $A\cap B=\{2,3,5\}\cap\{2,4,6\}=\{2\}$