Flyingabhi wrote:
ThatDudeKnows wrote:
There are six options for X, each of which has a probability of 1/6. Let's explore each of the options for X.
If X=1, there are six options for Y (2, 3, 4, 5, 6, or, 7), each of which has a probability of 1/6. So the probability of X=1 and Y=3 is 1/6*1/6 = 1/36.
If X=2, there are five options for Y (3, 4, 5, 6, or, 7), each of which has a probability of 1/5. So the probability of X=2 and Y=3 is 1/6*1/5 = 1/30.
If X>2, 3 won't be available as an option for Y, so the probability of X={3,4,5,6} and Y=3 is zero.
(X=1, Y=3) and (X=2, Y=3) are the only two ways to "win" and we know each of their probabilities. Let's add them together to find the total probability of winning.
1/36 + 1/30 = 5/180 + 6/180 = 11/180
Answer choice A.
I approached this question in a different way and could not get the answer mentioned in the options.
Total num of ways in which two integers can be selected:
If 1st person chooses 1, 2nd can choose out of 6 integers, if he chooses 2, then 2nd can choose amongst 5 and so on..
So total number of ways= (6+5+4+3+2+1)=21
Out of these outcomes only two are the desired outcomes.(1,3) and (2,3).
So probability of getting 3 is 2/21.
Where am i going wrong? Please help.
Posted from my mobile deviceYou are correct that there are 21 possible outcomes, but they are not equally likely. We have a CONDITIONAL PROBABILITY issue. Let's look at an analogous situation.
If I work from home, I always eat soup for lunch. If I go to the office, I choose randomly from among six options (sandwich, salad, hamburger, chicken, steak, and pasta), each with equal probability, at the restaurant next door. I work from home half the time. What percentage of the time do I eat soup for lunch? By your methodology, there are a total of seven options and soup is one of the options, so 1/7? Nope.
It should be clear that I have soup 1/2 the days.
You can only determine the probability of winning by doing
(count the ways to win) and divide by (count the total number of ways)
when all the ways have an equal probability. In my example, as well as in the question, the conditional probability component means that you need to calculate the probability of each individual way to win and then add up each of those probabilities.
Make sense?