Find probability in card question Suppose that a test for extrasensory perception consists
of naming (in any order) $3$ cards randomly drawn from
a deck of $13$ cards. Find the probability that by chance
alone, the person will correctly name (a) no cards, (b) $1$
card.
this is what I got
a) $f(0)= \left( \begin{array}{ccc}
3  \\
0  \\
 \end{array} \right)   (10/13)^3= 45.5$%
b)$f(1)= \left( \begin{array}{ccc}
3  \\
1  \\
 \end{array} \right) (3/13)  (10/13)^2=40.96$%
but the book says a) $42$% and b $47.2$%. I really need to understand this. So would anyone please show me where I am wrong.
 A: A) Assuming that each card is unique:


*

*The number of different ways to pick $3$ cards out of all $13$ cards is $\binom{13}{3}=286$

*The number of different ways to pick $3$ cards out of the $10$ cards not drawn is $\binom{10}{3}=120$

*So the probability that a person will name $0$ cards correctly is $\displaystyle\frac{120}{286}=41.9\%$



B) Assuming that each card is unique:


*

*The number of different ways to pick $3$ cards out of all $13$ cards is $\binom{13}{3}=286$

*The number of different ways to pick $2$ cards out of the $10$ cards not drawn is $\binom{10}{2}=45$

*The number of different ways to pick $1$ card out of the $3$ cards drawn is $\binom{3}{1}=3$

*So the probability that a person will name $1$ card correctly is $\displaystyle\frac{45\cdot3}{286}=47.2\%$

A: Assuming the cards are drawn without replacement and are each uniques, Barak's answer is correct for part (a). Diane's answers are right if the cards are drawn with replacement.
The book obviously meant the cards are drawn without replacement.
For part (b) the answer is 
$$
\binom{3}{1} \frac{3}{13} \frac{10}{12} \frac{9}{11} = \frac{135}{286}
$$
which is 47.2%. 
(Barak's method is superior but I gave a more step-by-step way to reach the same number because Barak already showed the couning-cases way.)
