You roll a pair of fair dice until you roll “doubles” (i. e., both dice are the same). what is the expected number, e[n], of rolls?

We need to define our outcomes and events.

Finding the probability of each event occurring separately, and then multiplying the probabilities is the step to finding  the probability of two independent events that occur in sequence.

To solve this problem, we take note of this:

The roll of the two dice are denoted by the pair

(I, j) ∈ S={ (1, 1),(1, 2),..., (6,6) }

Each pair is an outcome. There are 36 pairs and each has probability 1/36. The event “doubles” is { (1, 1),(2, 2)(6, 6) } has probability p= 6/36 = 1/6. If we define ”doubles” as a successful roll, the number of rolls N until we observe doubles is a geometric (p) random variable and has expected value E[N] = 1/p = 6.

your answer is a

step-by-step explanation:

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i believe it is c.y = 9x

rise over run. 9 units up over 1 unit right or 9/1