Probability doubt

Question

Hi guys,

I need to know the difference between these two problems. One of them is a link that you need to click.

Problem 1 
What is the probability that, on three rolls of a single fair die, ATlEAST ONE
of the rolls will be a six?

Problem 2 
 https://gmatclub.com/forum/dice-124986.html

Seems to me like both are the same and the same method is to be used. However, problem 1 does not take into account the same considerations in problem 2.

KarishmaB · Answer

The two problems are different.

Problem 1: You need to find the probability of getting something like this:
6, 6, 2 or
1, 6, 3 or
3, 4, 6 or
6, 6, 6 etc

Problem 2: You need to find the probability of getting something like this:
3, 2, 1 or
2, 1 or
5, 6, 1 or
4, 5, 1 or
1 etc

I hope you see the difference.
In the first problem, you can have the 6 anywhere and any number of times as long as you have at least one 6.
Probability of no 6s = (5/6)*(5/6)*(5/6)
So probability of at least one 6 = 1 - (5/6)*(5/6)*(5/6)

In the second problem, the moment you get a 1, you stop. You have at most 3 throws available. So there are 3 ways of doing it:
You get 1 on the first throw. Probability = 1/6
You get 1 on the second throw. Probability = (5/6)*(1/6)
You get 1 on the third throw. Probability = (5/6)*(5/6)*(1/6)
Add them up to get the required probability.

petrifiedbutstanding · Answer

Ah OK.. So in the first problem, the outcome of each throw does not influence the probability of the next throw because there is no special clause; whereas in the second problem, the probability is affected because of the clause that you will stop as soon as you derive the result.

Is my analogy correct?? I understand each person may visualize this a different way. So do better my explanation if wish to..

Is my analogy correct?? I understand each person may visualize this a different way. So do better my explanation if wish to..

KarishmaB · Answer

Yeah, that's fine.

petrifiedbutstanding · Answer

Thank you! I really like how you explained it..

vibhav · Answer

converting the seemingly difficult sums into a form such as 1-(probability of not getting a 6 in any throw) i.e. 1- (5/6*5/6*5/6) is really a great way of solving probability related Questions.

cyberjadugar · Answer

Hi,

A more generalized expression would be:
Probability of event happening at least once = 1 - (probability of non-occurrence of the event)

Regards,

Probability doubt

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