yasir9909 wrote:
Protia rates all her first dates as either "duds" or "dudes". Her date on Wednesday night was a dud.On the next night, she went on a date with someone else who was also a dud If the probability of her getting two duds in a row was 4/25,what is the probability that her next date will be a dude?
(A). 1/5
(B). 9/25
(C). 2/5
(D). 3/5
(E). 21/25
Although not stated, the assumption here is that, on ANY GIVEN DATE, the probability of getting a dud is always the same.
Let y = P(getting a dud on a given date)
We're told that P(2 duds in a row) = 4/25
In other words: P(dud on 1st date
AND dud on 2nd date) = 4/25
Rewrite as: P(dud on 1st date)
X P(dud on 2nd date) = 4/25
So, we get: y
X y = 4/25
In other words, y² = 4/25
Solve to get: y = 2/5
So, P(getting a dud on a given date) = 2/5
So, P(NEXT date is a dude) = 1 - 2/5 = 3/5
Answer:
I am ok with this explanation but since in the question is clearly specified that the second dud is different from the first one how can it be equally likely that the guy she dates is a dud? Assuming there is a certain number X of dud guys and N is the total number of guys, shouldn't this probability be \(\frac{X}{N}*\frac{X-1}{N-1}\). It is as there is no reimmision. Am I wrong?