**Given:**

An unbiased coin is tossed.

If heads occurs, a pair of dice is rolled and sum is noted.

If tails occurs, a number from 2, 3, 4,…, 12 is noted.

There are two mutually exclusive ways to note a number, which is either 7 or 8 –

**a.** Toss result is a head, and then, sum of the obtained numbers on rolling two dice is 7 or 8

**b.** Toss result is a tail, and then, a card numbered 7 or 8 is picked from the pack

Let E_{1} be the event that the result of the toss is a head and E_{2} be the event that the result of the toss is a tail.

Since there are only two outcomes for a coin toss and each outcome has an equal probability of occurring, we have

Let E_{3} denote the event that the noted number is 7 or 8.

Hence, we have

When two dice are rolled, we obtain a sum of 7 when the outcomes are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1).

We obtain a sum of 8 when the outcomes are (2, 6), (3, 5), (4, 4), (5, 3) and (6, 2).

Hence, 11 ways out of 36 give the sum 7 or 8.

Using the theorem of total probability, we get

P(E_{3}) = P(E_{1})P(E_{3}|E_{1}) + P(E_{2})P(E_{3}|E_{2})

Thus, the probability of the noted number being 7 or 8 is \(\cfrac{193}{792}\)