Suppose A and B are two events, not independent. Is the probability P(A and B) > 1/3?
Statement #1: P(A) = 0.8 and P(B) = 0.7
Statement #2: P(A or B) = 0.9
A
please i need someone to explain..
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
probability question
This topic has expert replies
-
Estrella84
- Newbie | Next Rank: 10 Posts
- Posts: 3
- Joined: Mon Nov 16, 2015 9:06 am
-
DavidG@VeritasPrep
- Legendary Member
- Posts: 2663
- Joined: Wed Jan 14, 2015 8:25 am
- Location: Boston, MA
- Thanked: 1153 times
- Followed by:128 members
- GMAT Score:770
Check out this post on probability from one of our instructors: https://www.veritasprep.com/blog/2016/01 ... tion-gmat/
-
Matt@VeritasPrep
- GMAT Instructor
- Posts: 2630
- Joined: Wed Sep 12, 2012 3:32 pm
- Location: East Bay all the way
- Thanked: 625 times
- Followed by:119 members
- GMAT Score:780
"Not independent" means that A and B are related in some way. For instance, suppose that I really like Vietnamese food, and that there's a 70% chance that I'm going to get Vietnamese food for lunch today.
Also suppose that IF I get Vietnamese food for lunch, I'll definitely (100%) get a durian smoothie, since the place in my town offers one and it's great. But I might also get a durian smoothie (20%) anyway, no matter where I get lunch, since I'm so used to have it for dessert.
As you can see, then, my probability of getting a durian smoothie is NOT independent of my probability of having Vietnamese food. There are two cases:
Case I: Vietnamese and Durian
In this case, we start with Vietnamese (70%), then multiply it by durian GIVEN THAT I'm getting Vietnamese (100%). So there's a 70% chance this happens.
Case II: No Vietnamese, but still Durian
In this case, we start with No Vietnamese (30%), then multiply it by durian GIVEN THAT I didn't get Vietnamese (20%). So there's a 6% chance that this happens.
Adding the two, my probability of getting a durian smoothie today is 76% (and that probability is rising as I type this!)
Now let's apply this to our problem.
Taking S1, let's assume that the probability of A and B happening is less than 1/3: say 20%.
That means that P(A and B) = .2. So P(A and NOT B) = .6, since the two together must add up to P(A). (Intuitively, this means that A = (A with B) + (A without B), which should make sense.)
But then we also have P(B) = .7, which implies that P(B and NOT A) = .5. All of a sudden we have too much probability: P(A and B) + P(A and NOT B) + P(B and NOT A) = 1.3, which is too high!
From this we can see that our minimum P(A and B) should come when we assume that one of A and B must happen, and we treat this like a Venn diagram. If we do P(A) + P(B) - P(A and B) = 1, we find that P(A and B) = 0.5.
The probability of P(A and B) can be higher, of course - as high as 0.7. But it can't be any higher than that, since P(A and B) can't be bigger than P(B), which itself is 0.7.
So 0.7 ≥ P(A and B) ≥ 0.5, and we're done!
S2 is less helpful: A and B could be mutually exclusive, for instance, and P(A and B) could equal 0. (For instance, suppose P(A) = the probability that the Seahawks win the Super Bowl (89%) and P(B) = the probability that the Patriots do (1%).)
Also suppose that IF I get Vietnamese food for lunch, I'll definitely (100%) get a durian smoothie, since the place in my town offers one and it's great. But I might also get a durian smoothie (20%) anyway, no matter where I get lunch, since I'm so used to have it for dessert.
As you can see, then, my probability of getting a durian smoothie is NOT independent of my probability of having Vietnamese food. There are two cases:
Case I: Vietnamese and Durian
In this case, we start with Vietnamese (70%), then multiply it by durian GIVEN THAT I'm getting Vietnamese (100%). So there's a 70% chance this happens.
Case II: No Vietnamese, but still Durian
In this case, we start with No Vietnamese (30%), then multiply it by durian GIVEN THAT I didn't get Vietnamese (20%). So there's a 6% chance that this happens.
Adding the two, my probability of getting a durian smoothie today is 76% (and that probability is rising as I type this!)
Now let's apply this to our problem.
Taking S1, let's assume that the probability of A and B happening is less than 1/3: say 20%.
That means that P(A and B) = .2. So P(A and NOT B) = .6, since the two together must add up to P(A). (Intuitively, this means that A = (A with B) + (A without B), which should make sense.)
But then we also have P(B) = .7, which implies that P(B and NOT A) = .5. All of a sudden we have too much probability: P(A and B) + P(A and NOT B) + P(B and NOT A) = 1.3, which is too high!
From this we can see that our minimum P(A and B) should come when we assume that one of A and B must happen, and we treat this like a Venn diagram. If we do P(A) + P(B) - P(A and B) = 1, we find that P(A and B) = 0.5.
The probability of P(A and B) can be higher, of course - as high as 0.7. But it can't be any higher than that, since P(A and B) can't be bigger than P(B), which itself is 0.7.
So 0.7 ≥ P(A and B) ≥ 0.5, and we're done!
S2 is less helpful: A and B could be mutually exclusive, for instance, and P(A and B) could equal 0. (For instance, suppose P(A) = the probability that the Seahawks win the Super Bowl (89%) and P(B) = the probability that the Patriots do (1%).)
