## anonymous one year ago John’s piggy bank contained 7 dimes and 3 quarters. He pulled out 1 coin without looking. Without replacing the first coin, John then pulled out a second coin. What is the probability that both coins were dimes? Express your answer in simplest form.

1. anonymous

My thought is that it's as simple as 2/10. Im not quite sure though.

2. anonymous

We have two steps for this. I'll help you through them one-by-one. The first step is figuring out what the probability is for pulling the first dime out. Which is represented as $\large \frac{no.~of~dimes}{total~no.~of~coins}$

3. anonymous

7/10 then I would presume?

4. anonymous

Yep. So our next step (3 in total, not 2 sorry) is to figure out what the probability of getting a dime would be if there is one less dime and one less coin. Which is represented as $\large \frac{no.~of~coins~-1}{total~no.~of~coins~-1}$

5. anonymous

So 6/9. With the reduced dime he would have presumably pulled, it takes one away from the amount of dimes, which in return reduces the amount of coins altogether.

6. anonymous

Right, good job. Before we move on to the final step, do you know what 6/9 will simplify to?

7. anonymous

2/3 of course.

8. anonymous

Great! So our final step is to multiply the two fractions together, to get the probability of both happening. So $\large \frac{7}{10} \times \frac{2}{3}=~?$

9. anonymous

14/30, which would reduce to 7/15. 7 is a prime number so that must be my answer.

10. anonymous

Cool! Thanks a lot. I appreciate your help!

11. anonymous

Right, it would be 7/15. And also, $\huge \color{aqua}N\color{fuchsia}o \space \color{lime}P \color{orange}r \color{blue}o \color{maroon}b \color{red}l \color{olive}e \color{purple}m \ddot\smile$