# An unbiased coin is tossed. If the result is a head, a pair of unbiased dice is rolled and the sum of the numbers obtained is noted.

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An unbiased coin is tossed. If the result is a head, a pair of unbiased dice is rolled and the sum of the numbers obtained is noted. If the result is a tail, a card from a well shuffled pack of eleven cards numbered 2,3,4,…,12 is picked and the number on the card is noted. What is the probability that the noted number is either 7 or 8?

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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 E1 be the event that the result of the toss is a head and E2 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 E3 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(E3) = P(E1)P(E3|E1) + P(E2)P(E3|E2) Thus, the probability of the noted number being 7 or 8 is $\cfrac{193}{792}$