Probability that leftmost lamp is not chosen and rightmost lamp is chosen

Question

Ryan has 3 red lava lamps and 3 blue lava lamps. He arranges them in a row on a shelf randomly, and then randomly turns 3 of them on. What is the probability that the leftmost lamp is blue and off, and the rightmost lamp is red and on?

Here's my thought process. There are $\frac{6!}{3!3!}$ ways to arrange the lamps. Of those $20$ ways, there are $\binom{6}{3}$ ways to turn the lamps on. This gives us $400$ different possibilities. From here, I'm not sure what to do. Help is greatly appreciated.

Answer 1

Let $B$ denote the event that the leftmost lamp is blue and let $R$ denote the event that the rightmost lamp is red.

Then: $$P(B\cap R)=P(B)P(R\mid B)=\frac12\frac35=0.3$$

On sortlike way you can compute that there is a chance of $0.3$ that the leftmost lamp is off and the rightmost lamp is on.

There is independence so that the probability of the event mentioned in your question is: $$0.3\times0.3=0.09$$

Answer 2

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?