Probability (Help)

[hoenhoen2211]

A fair pair of dice is rolled until the sum of the two numbers occurs as 7. Compute the probability that an even number of rolls are needed.

can someone help me with this? Thank you

I have no idea what to do with this exercise…

 

[BigBeachBanana]

A fair pair of dice is rolled until the sum of the two numbers occurs as 7. Compute the probability that an even number of rolls are needed.

can someone help me with this? Thank you

I have no idea what to do with this exercise…

I would start listing out the exhaustive list of 2 possible rolls that add up to 7.

 

[JeffM]

I would start by computing three probabilities

7 is achieved on the first roll
7 is not achieved on the first roll but is achieved on the second roll
7 is not achieved on either the first or second roll

Now how would you compute the probability that exactly 2 or exactly 4 rolls are needed?

 

[pka]

A fair pair of dice is rolled until the sum of the two numbers occurs as 7. Compute the probability that an even number of rolls are needed.

Look HERE, In that expansion the term [imath]{\large 6x^7}[/imath] tells us that the sum of seven is obtained in six ways.
Thus the probability of a sum of seven on any given toss is [imath]\dfrac{6 }{36}=\dfrac{1}{6}[/imath].
So the probability of the first six on the nth toss is [imath]\left(\dfrac{5}{6}\right)^{n-1}\left(\dfrac{1}{6}\right)[/imath].
What do you know about [imath]\large\bf n~?[/imath] Can you sum a series?


[imath][/imath][imath][/imath]

 

[pka]

Look HERE, In that expansion the term [imath]{\large 6x^7}[/imath] tells us that the sum of seven is obtained in six ways.
Thus the probability of a sum of seven on any given toss is [imath]\dfrac{6 }{36}=\dfrac{1}{6}[/imath].
So the probability of the first six on the nth toss is [imath]\left(\dfrac{5}{6}\right)^{n-1}\left(\dfrac{1}{6}\right)[/imath].
What do you know about [imath]\large\bf n~?[/imath] Can you sum a series?


[imath][/imath][imath][/imath]

Correction: So the probability of the first seven on the nth toss is [imath]\left(\dfrac{5}{6}\right)^{n-1}\left(\dfrac{1}{6}\right)[/imath].

 

[hoenhoen2211]

Correction: So the probability of the first seven on the nth toss is [imath]\left(\dfrac{5}{6}\right)^{n-1}\left(\dfrac{1}{6}\right)[/imath].

Thank you!!!