<s>[INST] What is the probability of getting 2 same colour sweets and 1 different colour sweet? A little box contains $40$ smarties: $16$ yellow, $14$ red and $10$ orange.
You draw $3$ smarties at random (without replacement) from the box.
What is the probability (in percentage) that you get $2$ smarties of one color and another smarties of a different color?
Round your answer to the nearest integer.
Answer given is $67$. I don't get it. Is it not:
$$\left(\frac{16}{40} \times \frac{15}{39} \times\frac{24}{38}\right) + \left(\frac{14}{40} \times\frac{13}{39} \times\frac{26}{38}\right) +\left(\frac{10}{40} \times\frac{9}{39} \times\frac{30}{38}\right)= 22?$$
 [/INST] Your options are "Exactly two yellow smarties, exactly two red smarties, or exactly two orange smarties."
If $P$ represents your final probability, you need to add up the following probabilities:
$P($exactly two yellows$) \, + \, P($exactly two reds$) \, + \,P($exactly two orange$)$
The probability of getting exactly two yellow smarties is $3\cdot \frac{16}{40}\cdot \frac{15}{39}\cdot \frac{24}{38}$. The reason we multiply by $3$ is that there are three different ways to choose two yellow smarties, i.e. $\binom{3}{2}=3$. The first two draws can be yellow, the first and the last can be yellow, or the last two draws can be yellow.
Similarly we can find $P($exactly two reds$)$ and $P($exactly two orange$)$. 
Thus $P=3\left( \frac{16}{40}\cdot \frac{15}{39}\cdot \frac{24}{38}\right)+ 3\left( \frac{14}{40}\cdot \frac{13}{39}\cdot \frac{26}{38}\right)+3\left( \frac{10}{40}\cdot \frac{9}{39}\cdot \frac{30}{38}\right)\approx 67$%.
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