Probability problem... Fiza has 10 coins in a bag
There are three £1 coins and seven 50p coins

Fiza takes at random, 3 coins from the bag

Work out the probability that she takes exactly £2.50

I found this problem on a practice paper my teacher gave me. Initially, I thought I'd have to do a  tree diagram outlining her first pick, second pick and third pick. But then it doesn't say that she takes them out separately, meaning she must have taken the three coins at once (and also because I thought I'd be over complicating things when not necessary).
I got an answer of 63/1000, however I just wanted to check from the experts and also because this question was worth 4 marks in an actual exam, but my working out (3/10 * 3/10 * 7/10) would say otherwise.
Note: If I am wrong please specify how to get the right answer
 A: It is actually a little more intricate than what you've done. You can describe each pick individually, but you then have to count each branching point individually:
First pick: she can either take £1 or 50p with probabilities $3/10$ and $7/10$ respectively
Second pick: if she took 50p at first then she'll now need to get £1 twice to sum 2.50
Third pick: if the coins sum to £1.50, pick £1, otherwise, pick 50p
Also, once she has made her pick, for the second coin you have to consider you have 1 coin less to choose from.
This gives: $\frac{3}{10}\frac{2}{9}\frac{7}{8}+ \frac{3}{10}\frac{7}{9}\frac{2}{8} + \frac{7}{10}\frac{3}{9}\frac{2}{8} = \frac{7}{40}$ 
A: Find the number of ways of drawing 3 coins and getting 2.50, and divide it by the total number of ways of drawing 3 coins. To get 2.50, Fiza must have drawn any two 1-pound coins and any 1 half-pound coin.There are 3 ways to draw 2 1-pound coins, and 7 ways to draw one half-pound coin. So, $3\times 7$ is the numerator.
To get the denominator, we need combination of 10 coins to 3 so it is 
$\frac{10!}{((10!-3!)\times (3!))} = 120$.
So the answer is $\frac{(3\times7)}{120} = \frac {7}{40}$.
A: With a little combinatorial set theory:
$$\frac{3\cdot7}{\binom{10}3}=\frac{21}{120}=\frac{7}{40}\;,\;\;\text{with}$$
$$3\cdot 7=21=\;\text{ways to choose two 1 pound coins ($\,3\,$) and one $\;50\;$ penny coin ($\,7\,$),}$$
or number of subsets with three elements two of which are taken from a subset with three elements (the 1 pound coins) and the other seven elements from the subset of 50 p. coins
$$\binom{10}3=\;\text{number of ways to choose three coins out of a set with ten coins,}$$
or number of subsets with three elements that a set with ten elements has.
A: In order to get £2.50 with only 3 coins then Fiza has to draw two pound coins and a 50p piece.  There are thus three ways Fiza can do this as the 50p could be drawn first, second or last.
So the probability is
$$3 \times\frac{3}{10}\cdot\frac{2}{9}\cdot\frac{7}{8} = \frac{126}{720} = \frac{7}{40}$$
Here I'm assuming you draw the 50p last so for the first draw there are 10 coins in the bag 3 of which are £1 the probability of the first coin drawn being a pound is $\frac{3}{10}$.  Now there are 9 coins in the bag of which two are one pound so the probability of the second coin being a pound is $\frac{2}{9}$ and for the final coin there are now 8 coins in the bag and 7 of them are 50p so the probability is $\frac{7}{8}$.  We multiply this by 3 because the 7 could appear in three places depending on whether the 50p comes first, second or last. 
A: As the number of coins and the amount to be taken from the bag are both conveniently less we can figure out all possible combinations of $1$ ruppee and $50$ paisa coins Fiza can take out to make a total of $2.50$ ruppees:


*

*2 one ruppee coins, 1 fifty paise coin

*1 one ruppee coin, 3 fifty paise coins

*5 fifty paise coins


The last two combinations are not feasible, since Fiza takes out only 3 coins. The probability of the first combination is $\frac{3}{10}.\frac{2}{9}.\frac{7}{8}\times 3$ 
A: Unfortunately it's wrong. For discrete objects probability can be thought of as the number of solutions (desired configurations) over the number of (equally probable) possibilities. In this case what are the possibilities? She cannot take out the same coin as 3 coins, so you have to take that into account. There are 10 coins, so there are $\binom{10}{3}$ ways to take out 3 coins. Likewise, the two dollar coins in a solution must be different coins out of 3, so there are $\binom{3}{2}$ ways for that.
