aggvipul wrote:
arosman wrote:
aggvipul wrote:
Q. Given that there are 5 married couples. If we select only 3 people out of the 10, what is the probability that none of them are married to each other?
I want to solve this question using reverse probability approach, can someone post the solution pls.
OA - \(\frac{2}{3}\)
Source:
https://gmatclub.com/forum/math-probability-87244.htmlI don't think using the reverse probability(probability none are married = 1 - probabilty at least one pair is married) is the best approach. The reason for this is because we have to take into account the fact that there are multiple ways of getting a couple ---> pick 1 is married to pick 2, pick 2 is married to pick 3, pick 1 is married to pick 3.
Probability no couples = (Number of 3 person selections with no couples)/Total number of selections of 3 people from 10
Number of 3 person selections with no couples = (10 * 8 * 6)/3! ---->80
-Note: We have to divide by 3! to get rid of repeats(ie ABC is the same team as BAC)
Total number of selections of 3 people from 10 ----> 10C3 = 10!/(7! * 3!) = 120
80/120 = 2/3
Hi
arosman while I agree that reverse probability approach may not be the best here but as mentioned in my post the purpose to get the answer through reverse probability is to learn the concept more thoroughly.
bb can you pls help out here
Start at highest level possible.
Probability no couples picked + probability at least one couple picked = 1
Therefore -----> probability no couples picked = 1 - probability at least one couple picked
Since only three people are picked total, there is no way to select more than one couple. Therefore ---> probability at least one couple picked = probability exactly one couple picked
Probability exactly one couple picked = probability picks 1 & 2 are a couple + probability picks 1 & 3 are a couple + probability picks 2 & 3 are a couple
Since all 3 of these options are going to have the same probability we can just think of this as ---> probability exactly one couple picked = 3 * probability picks 1 & 2 are a couple
Imagine we have 3 slots.
Slot 1: Slot 1 can be any person as long as slot two is that person's partner. Probability of selecting any person is just 10/10 = 1
Slot 2: There are nine people left but only 1 of them is Slot 1's partner. Therefore probability here is 1/9
Slot 3: There are 8 people left and since we already have out couple any 8 of them can be selected. 8/8 = 1
Therefore, the exact probability of this exact sequence(Partner 1 ---> Partner 2 ---> Anybody else) is 1 * 1/9 * 1 = 1/9.
Accounting for the 3 different scenarios ------> Probability at least one couple = 3 * 1/9 = 1/3
Probability no couples = 1 - probability at least one couple = 1 - 1/3 = 2/3
Lemme know if you have any questions!
Best,
Adam
Can this approach be taken if I want to solve it taking couples (for the learning purpose) using reverse probability. I am using the below methodology but the answer is not correct
Prob of choosing 3rd person from a different couple = \(\frac{4}{4}∗\frac{1}{2}\)
Hence, Prob. =\(\frac{5}{5}∗\frac{2}{2}∗\frac{4}{4}∗\frac{1}{2}∗2\) = 1