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 1

The probability that the leftmost lamp on the shelf is red and the leftmost lamp which is turned on is also red is 1/240.

Step-by-step explanation:

The number of ways to arrange the 6 lava lamps on the shelf can be calculated by using the concept of permutations. There are a total of 6 lamps, consisting of 3 red and 3 blue lamps. So, the total number of arrangements is 6! = 720.

Now, to find the probability that the leftmost lamp on the shelf is red and the leftmost lamp which is turned on is also red, we need to consider the favorable outcomes. If the leftmost lamp is red, it means one of the 3 red lamps must be in the leftmost position. For the leftmost lamp that is turned on to be red, it means out of the 3 red lamps, 1 of them must be turned on.

Therefore, the probability can be calculated as:

(Number of favorable outcomes) / (Total number of possible outcomes) = (3 * 1) / 720 = 1/240

Answer 2

There are 6! = 720 ways of arranging the lamps.

If the leftmost lamp is red, there are 3 choices of lamp in the leftmost position, and the remaining 5 can be placed in any order, so there are 3×5! = 360 ways of arranging the lamps and the leftmost is red.

Hence there is a 360/720 = 1/2 probability that the leftmost lamp is red.

Ignoring lamp color for the moment, the probability of arranging 3 lit lamps and 3 unlit lamps is the same, 1/2.

Since Ryan arranges the lamps randomly by color, then turns 3 of them on randomly, the two events are independent. So

P(leftmost red AND leftmost lit) = P(red) × P(lit) = 1/2² = 1/4