probability in a fair die

[amal_]

If a die is thrown repeatedly ,what is the probability that a 6 is obtained before a 3 turns up.

 

[pka]

If a die is thrown repeatedly ,what is the probability that a 6 is obtained before a 3 turns up.

Suppose that [imath]X[/imath] is the number of the toss upon which the first six appears and is such that for any [imath]T<X[/imath] on toss [imath]T[/imath] a three does not appear.
Can you explain each of these? [imath]\mathcal{P}(X=1)=\dfrac{1}{6}[/imath], [imath]\mathcal{P}(X=2)=\left(\dfrac{1}{6}\right)\left(\dfrac{4}{6}\right)[/imath], and [imath]\mathcal{P}(2<X)=\left(\dfrac{1}{6}\right)\left(\dfrac{4}{6}\right)^{X-1}[/imath].
Please reply with your answers.

 

[amal_]

P(X=1)=1/6 means that that we tossed the die once and and we obtained a 6 .

P(X=2)=(1/6)*(4/6) means that we tossed the die twice ,the first time we obtained any number other than 3 and 6,the second time we obtained a 6.

P(2<X)=(1/6)*(4/6)^(X−1) means that we tossed the die X times :For the first (X-1)th time a 3 and 6 didn't appear but for the X th time a six appeared for the first time

Would that be correct ?

 

[pka]

P(X=1)=1/6 means that that we tossed the die once and and we obtained a 6 .
P(X=2)=(1/6)*(4/6) means that we tossed the die twice ,the first time we obtained any number other than 3 and 6,the second time we obtained a 6.
P(2<X)=(1/6)*(4/6)^(X−1) means that we tossed the die X times :For the first (X-1)th time a 3 and 6 didn't appear but for the X th time a six appeared for the first time
Would that be correct ?

Well done, yes they are correct. Can you find [imath]\sum\limits_{k = 1}^\infty {\left( {\frac{1}{6}} \right){{\left( {\frac{4}{6}} \right)}^{k - 1}}}~?[/imath] And what is the point?

 

[Steven G]

Is not the answer obvious??

 

[amal_]

Well done, yes they are correct. Can you find [imath]\sum\limits_{k = 1}^\infty {\left( {\frac{1}{6}} \right){{\left( {\frac{4}{6}} \right)}^{k - 1}}}~?[/imath] And what is the point?

That would be the sum of the probabilities of the event 6 appears for the first time the k th time and no six or three appear any of the (k-1)th times ,hence the probability that a 6 is obtained before a 3 turns up.

It is equal to 1/2.

I am still learning the basics of probability so this has been extremely helpful. Thank you so much for your help !

 

[Steven G]

There is a 50-50 chance that a 6 will be rolled before a 3. This should have been clear from the start. That is, the probability that a 6 occurs before a 3 is 1/2.

 

[Dr.Peterson]

There is a 50-50 chance that a 6 will be rolled before a 3. This should have been clear from the start. That is, the probability that a 6 occurs before a 3 is 1/2.

This is only obvious if you happen to look at it from the right perspective -- which is probably not the perspective most students will come to the question with, expecting to use standard methods like pka's.

You have to back off from it and realize that the answer will be identical if you change it to "what is the probability that a 3 is obtained before a 6 turns up?"

Symmetry arguments rarely apply, but when they do, they are powerful. (On the other hand, more complicated arguments then prove themselves when they agree with the "trick" method.)