Question:
Ryan has 3 red lava lamps and 3 blue lava lamps. He arranges them in a row on a shelf randomly, then turns 3 random lamps on. What is the probability that the leftmost lamp on the shelf is red, and the leftmost lamp which is turned on is also red?

Answer:
There are $\binom{6}{3}=20$ ways for Ryan to arrange the lamps, and $\binom{6}{3}=20$ ways for him to choose which lamps are on, giving $20\cdot20=400$ total possible outcomes. There are two cases for the desired outcomes: either the left lamp is on, or it isn't. If the left lamp is on, there are $\binom{5}{2}=10$ ways to choose which other lamps are on, and $\binom{5}{2}=10$ ways to choose which other lamps are red. This gives $10\cdot10=100$ possibilities. If the first lamp isn't on, there are $\binom{5}{3}=10$ ways to choose which lamps are on, and since both the leftmost lamp and the leftmost lit lamp must be red, there are $\binom{4}{1}=4$ ways to choose which other lamp is red. This case gives 40 valid possibilities, for a total of 140 valid arrangements out of 400. Therefore, the probability is $\dfrac{140}{400}=\boxed{\dfrac{7}{20}}$.