Question 1096884
<br>I will assume since you are working a problem like this that you are familiar with "n choose r", which I will represent with C(n,r).  That number is the number of ways of choosing r of n identical objects.<br>
Overall, you are choosing 10 of the total 25 cars for the display, so the total number of possible choices (the denominator of your probability fraction) is C(25,10).<br>
For case (a), you need to choose 7 of the 7 red cars, AND 3 of the 14 silver cars, AND 0 of the 4 black cars.  As always when calculating probabilities, those "AND"s mean you are going to multiply the individual probabilities.  So for case (a), the numerator is [C(7,7)*C(14,3)*C(4,0)]<br>
Answer for part (a): {{{(C(7,7)*C(14,3)*C(4,0))/C(25,10)}}}<br>
By similar analysis, the answer for part (b) is {{{(C(7,6)*C(14,0)*C(4,4))/C(25,10)}}}<br>
For part (c) we need to do a bit more work.  We know all 7 red cars must be chosen (C(7,7)); but we could have either 3 silver and 0 black, OR 2 silver and 1 black, OR 1 silver and 2 black, OR 0 silver and 3 black.  And again, as always in probability those "OR"s mean we will be adding probabilities.<br>
So the numerator for case (c) is C(7,7) multiplied by the SUM of C(14,3)*C(4,0), C(14,2)*C(4,1), C(14,1)*C(4,2), and C(14,0)*C(4,3).<br>
Answer for part (c):
{{{(C(7,7)*(C(14,3)*C(4,0) + C(14,2)*C(4,1) + C(14,1)*C(4,2) + C(14,0)*C(4,3)))/C(25,10)}}}