Say you have 8 "cards". On each card there is a letter. One card has a G, one has an O, one has an M, one has a T, one has an R, one has a Y, and two have E's. Say you randomly select two cards, replacing the first before you select the second. Find the probability you select two E's.

Question

Say you have 8 "cards".  On each card there is a letter.  One card has a G, one has an O, one has an M, one has a T, one has an R, one has a Y, and two have E's.  Say you randomly select two cards, replacing the first before you select the second.  Find the probability you select two E's.

anonymous · Answer

So this is an independent probability problem.  The events do not rely on either outcome.  So the multiplication is rather simple.  You just need to write out the probability of each event and multiply them.  The first even could be defined as$$\frac{ 2 }{ 8 }$$ because out of the eight total cards, you want to draw one of the "e"s.  Now the second event is written the same because you simply replace the previously drawn card. 

Now you simply multiply $$\frac{ 2 }{ 8 }*\frac{ 2 }{ 8 }$$
Which comes to$$\frac{ 4 }{ 64 }$$
And reduces to $$\frac{ 1 }{ 16 }$$