Is the probability of Event A greater than 0.5

Dear Janvisahu,

I'm happy to respond.

My friend, I'm sorry, but your reasoning is flawed. Basically, you were assuming that Statement #2 was already true as you were reasoning with Statement #1.

Let's be clear: a formula-based approach to probability is doomed to failure. All the formulas in probability are context-specific, and if you don't understand exactly when you can and when you can't use one, then are driving blind through the realm.

The formula P(A and B) = P(A)*P(B) is ONLY true in the very rare and special case events A & B are independent. Most things in the real world are not independent, and if A & B are not independent, then that formula is entirely out-the-window.

For two events to be independent, truly independent, means that if we know the outcome of event A, that gives us absolutely no information about the outcome of event B, and vice versa. The following are real-world event pairs that are not independent
1) A = person earns over $200K annually, B = person has a college-degree or higher
2) A = a person eats pork, B = a person is a Muslim
3) A = a person has smoked for years, B = a person gets cancer
None of these pairs are independent, because if we know the answer to one question, that answer automatically makes the other event either more likely or less likely.

As it turns on, typically only artificial inanimate things are truly independent, such as the dots showing on two different dice rolled together, or two successful flips of a coin. A few very unusual rare things in the real world would be truly independent. For example, if A = a person is left-handed, and B = a person gets cancer--so far as I know, those two are entirely independent, because knowing the answer to one question leaves us totally in the dark about the answer to the other question.

Thus, most real events are NOT independent, and so the P(A and B) = P(A)*P(B) is useless and 100% inapplicable most of the time. We can only use that in the very special case in which A & B are guaranteed to be independent. Under statement #1, the two events could be independent, but they don't have to be independent.

Back to statement #1:
St-1: The probability that events A and B both occur is 0.6
Case (a): P(A) = 0 and P(B) = 0.6, so P(A and B) = 0.6. This is consistent with Statement #1 gives a prompt answer of no.
Case (b): P(A) = 0.8, P(B) = 0.75, events & A & B are independent: then P(A and B) = 0.6 This is consistent with Statement #1 gives a prompt answer of yes.
Consistent with Statement #1, we can construct scenarios that produce either a yes or no answer to the prompt. Thus, statement #1, alone and by itself, is not sufficient.

Statement #2 is not sufficient by itself either, but when we combine the statement, then we get the reasoning that you provided. Together, the statements are sufficient, and the correct answer is (C).

Does all this make sense?
Mike

mikemcgarry : Thanks for the wonderful explanation of the independent events. Kudos!

Thanks for your clarification, but I am a bit confused...
As per your explanation

""Back to statement #1:
St-1: The probability that events A and B both occur is 0.6
Case (a): P(A) = 0 and P(B) = 0.6, so P(A and B) = 0.6. This is consistent with Statement #1 gives a prompt answer of no. ""

If P(A) =0, it means A is impossible event, then how probability of both A and B can be non zero.

Please explain.

Hi..

You are correct.
If the probability of an event A is 0.x, the probability of it, A, happening with another event B will be LESS than 0.x
and equal to 0.x if the other event B is 100% to happen
this has to be true irrespective of type of event..




mikemcgarry, it seems you have taken OR in case (a) as the P(A and B) will be 0 in this case

independent events
P(a and b) = P(a)*P(b) so 0.6 = 0.5*P(b).. P(b)=1.2 ... cannot be >1..... NOT possible
dependent events
P(a and b) = P(a)*P(a/b) so 0.6 = 0.5*P(a/b).. P(a/b)=1.2 cannot be >1 NOT possible
P(a/b) is probability of b happening when a has already happened so it has to be less than P(a)

statement I is sufficient

Thank you for clarification of doubt. Kudos to you.
Please provide kudo, I helped in rectification of OA.

Bunuel, Gnpth

Please revise the OA to A.
As discussed in the post above, OA should be A.

Hi Janvisahu & chetan2u

In my opinion I would not consider statement 1 as sufficient for the simple reason that we do not know about the dependence and/or independence of events A & B
Now as you have taken into account that even for dependent events P(A)>0.5 but does this statement clarifies what is event A and what is event B and what is there dependency

Baye's theorem for conditional probability states that -

\(P(A and B) = P(A|B)*P(B)=P(B|A)*P(A) =>P(A|B)=\frac{P(B|A)*P(A)}{P(B)}\)

so per this statement and using your reasoning to find P(A), I can also have P(B)=0.6/P(A|B). Now P(B)>0.5 and we know nothing about P(A) here.

I'll illustrate this with simple example -

Q) If in a town population of 180,000 females, 66.66666% can expect to live to age 60, while 40% can expect to live to age 80. Given that a woman is 60, what is the probability that she lives to age 80?

This is an example of a conditional probability. In this case, the original sample space can be thought of as a set of 180,000 females. The events B and A are the
subsets of the sample space consisting of all women who live at least 60 years, and at least 80 years, respectively. We consider B to be the new sample space, and note
that A is a subset of B. Thus, the size of B is 120,000, and the size of A is 72,000.

So, the probability in question equals 72,000/120,000= 0.6. Thus, a woman who is 60 has a 60% chance of living to age 80

Clearly I have utilized events A & B in a way to show P(A)<0.5, here

The point I am trying to make is until and unless A & B are clearly defined we cannot use Statement 1 only to find the P(A). it can very well be used for P(B). Hence this statement is not sufficient.

I might be wrong here and if we have any expert on Baye's theorem/discrete conditional probability, then I would definitely like to learn few things and stand corrected.

I am also not sure whether Gnpth had thought of these scenarios before drafting the question. So we should give him the benefit of doubt and move ahead with Option C

Hi...

We have had a long discussion already...
Be it P(A) or P(B), Bayer's theorem or logic, none can be <0.6..



Let's look the way you have suggested..
P(A and B)=\(P(B)*P(\frac{A}{B})=0.6\)..
Take the max value of P(B) to get the least value of \(P(\frac{A}{B})\)..
\(1*P=0.6\)...so \(P(\frac{A}{B})=0.6\)..
That is prob of A happening given B has already happened and B is 100% so P(A) boils down to 0.6
If \(P(B)=0.6\), then \(P(\frac{A}{B})\)=1.. means if b is happening, a will also happen, so P(A) also becomes 0.6..

Now let's discuss on the example given by you..



You have taken P(60) as 1 and looking for P(80), so generally not the relationship we are looking for..

Also you have reduced the event to ONLY A, because you are taking people xg 80 as % of people xg 60 BUT the people xg 80 have to be taken as part of total..
The Q of yours should look for P(people xg 60 AND people xg 80).
Ofcourse it will be EQUAL to just people xg 80, so answer 40%

Both p(60 and 80) and p(80) or whatever you want to call is 0.4 and the other P(60) is 1
Thus it doesn't fit in here..

Hope it helps..

Okay. I haven't fully thought through it. But I did come across many problems which used the reasoning to arrive that we need both to solve.

But based on Chetan's analysis- I guess I will take look at some more examples to conclusively get the discrepancies out.

P.S- I didn't intend to make it debatable OA. But was sure about the approach in the examples I saw and my own reasoning.

Dear chetan2u,

It seems as if the debate is settling down at this point, but just for the record, I want to say to chetan2u: my brilliant colleague, you are 100% right and I made a pig-headed mistake.

Of course, if P(A and B) = 0.6, then for that 60% of the time, event A is certainly happening. Therefore, form Statement #1, we absolutely know that event A happens at least 60% of the time, no formula needed. Statement #1 is sufficient.

I don't know why I didn't think about it this way earlier. Thank you for pointing out my error, my friend.

Mike