Question 74918
1)
A) The probability of selecting one $5 bill is the number of $5 bills out of the total number of bills. So you have 4 $5 bills out of a total of 14. So this looks like:
{{{P(Selecting_5)=4/14=2/7}}}
So the probability of getting two 5's are
{{{P(5_and_5)=(4/14)*(4/14)=8/196=2/49}}} remember "and" means multiply the probabilities
So the chances of drawing 2 5's with the first one replaced is {{{2/49}}} or 4%


B)If you don't replace the bill taken out, then the total decreases by 1 for the 2nd probability
{{{P(5_and_5)=(4/14)*(4/13)=8/182=4/91}}}
So the chances of drawing 2 5's without the first one replaced is {{{4/91}}} or 4.3%


2)On a six-sided die there are 3 odd numbers out of 6 numbers. So the probability of rolling an odd number for the blue die is:
{{{P(odd)=3/6=1/2}}}
Since there are 2 numbers that are a multiple of 3 (3 and 6),the probability of rolling a multiple of 3 is:
{{{P(multiple_of_3)=2/6=1/3}}}
So the probability of these events occurring together are:
{{{P(odd_and_multiple)=(1/2)(1/3)=1/6}}}
It turns out that there is only one number that meets this condition. The number 3 is a multiple of 3 and is odd. Since the number 3 occurs once out of 6 numbers , the odds of rolling a 3 is 1/6, which justifies our answer. 



3)For the 1st digit you have the numbers 0-9 to choose from. So you have 10 numbers to choose from for the 1st digit. For the 2nd digit its the same, since you can repeat the digits. This is the same for all of the digits, so you get
{{{10*10*10*10=10^4=10000}}}
So there are a total of 10,000 combinations. It's easier to visualize this with a smaller set. Say we have 2 digits with only 0 and 1 as possible numbers, we would get this:
00,01,10,11
You can see there are 4 possible combinations. Since there are 2 numbers for both digits we can say
{{{2*2=4}}}
Which is our number of possible combinations. I hope that clears up things a bit. Currently I'm tutoring one-on-one for free. So feel free to send me any more problems or ask any more questions.