Understanding Geometric Distribution

[rclakmal]

Geometric Distribution?

Geometric Distribution: In an experiment, a die is rolled repeatedly until all six faces have finally shown.?
What is the probability that it only takes six rolls for this event to occur? ANSWER = 0.0007716

What is the expected waiting time for this event to occur? ANSWER = 35 rolls

Thanks a lot for any help.

 

[tiny-tim]

Geometric Distribution: In an experiment, a die is rolled repeatedly until all six faces have finally shown.?
What is the probability that it only takes six rolls for this event to occur? ANSWER = 0.0007716

What is the expected waiting time for this event to occur? ANSWER = 35 rolls

Thanks a lot for any help.

Show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[rclakmal]

yr sure ...i know that geometric distribution has the probability of (p)*(1-p)^(x-1) where the first success happens in x trials .

then i thought that ,
number to come in first trial is =(1/6)*(5/6)^(1-1)=1/6
number coming in the second trial =(1/6)*(5/6)^(2-1)=1/6*(5/6)
like wise

probability will be =sigma(x goes 1 to 6 )[(1/6)(5/6)^(x-1)]...but seems that i got the argument wrong.and can u please help me on this and if u can guide me for a site where some easy examples on this it will be really helpful.

 

[tiny-tim]

yr sure ...i know that geometric distribution has the probability of (p)*(1-p)^(x-1) where the first success happens in x trials .

That formula is really for where you're interested in a fixed result each time (for example, how many 4s are there?) …

Hint: here, you need the 2nd to be different from the 1st, the 3rd to be different from both of them, and so on

(btw, how did you get on with that other thread?)

 

[rclakmal]

yr then it should be 1*(5/6)*(4/6)*(3/6)*(2/6)*(1*6) is that so each time u have to remove one number ...so it should be like this no ...but I am not getting the answer., may be the answer is wrong ?or is it right .did u check it ?anyway thanks for ur helping hands


(what is other thread?i didnt get it ...!)

 

[tiny-tim]

yr then it should be 1*(5/6)*(4/6)*(3/6)*(2/6)*(1*6) is that so each time u have to remove one number ...so it should be like this no ...but I am not getting the answer., may be the answer is wrong

Yes, that's what I get.

The answer given, 0.0007716, is 1/6⁴ … I don't know how they get that.

(i meant https://www.physicsforums.com/showthread.php?t=311671)