0
$\begingroup$

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)=?

Answer:

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

$\endgroup$
  • $\begingroup$ $P(A\cap B)=P(A)P(B)$ is not joint-probability formula, it's independence formula. Of course, $A$ and $B$ are not independent here $\endgroup$ – user160738 Mar 6 '18 at 9:23
  • $\begingroup$ I found the formula from below source: investopedia.com/terms/j/jointprobability.asp $\endgroup$ – muhammad tayyab Mar 6 '18 at 9:26
  • $\begingroup$ 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. $\endgroup$ – muhammad tayyab Mar 6 '18 at 9:27
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.