An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the

Q: Why do we use the Law of Total Probability here? Can't we just add the probabilities?

Great question! You cannot simply add the probabilities because the two scenarios (rolling dice vs. picking a card) are not equally likely to occur in the same way. The Law of Total Probability acts as a weighted average. It accounts for the fact that you only enter the 'dice' scenario 50% of the time and the 'card' scenario the other 50% of the time. You must weight the conditional outcomes by the probability of the coin toss itself.

Q: How do I know if I've counted all the dice combinations correctly?

The best way to be sure is to be systematic. For a sum of 7, start with the smallest possible value for the first die (1) and see what the second die must be (6). Then increment the first die (2) and find the second (5), and so on. If you do this for all possibilities (1 to 6), you will never miss a pair. For a sum of 8, start with the smallest possible value for the first die that can result in a sum of 8 (which is 2, since 1+7 is impossible) and proceed from there.

Q: Why is the denominator 36 for the dice but 9 for the cards?

The denominator represents the total number of possible outcomes in that specific branch. When you roll two dice, each die has 6 faces, so the total outcomes are $6 \times 6 = 36$. When you pick one card from a deck of 9, there are only 9 possible outcomes. Always define your sample space based on the specific action being performed in that branch of the probability tree.

Question

An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noted. If the toss of the coin results in tail then a card from a well-shuffled pack of nine cards numbered 1,2,3,...,9 is randomly picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is :

Options:

(A) $\frac{13}{36}$

(B) $\frac{19}{36}$

(C) $\frac{19}{72}$

(D) $\frac{15}{72}$

Sigma Percentile · Accepted Answer

### The Crossroads of Chance: A Journey Through Probability Trees

Imagine you are standing at a fork in the road of a grand experiment. To your left, a path leads to the rhythmic clatter of dice; to your right, a path leads to the silent, shuffling anticipation of a card deck.

This is not just a math problem; it is a story of two parallel universes, and our job is to calculate the likelihood of a specific outcome—landing on a 7 or an 8—across both.

### Phase 1: The Fork in the Road

Every great probability journey begins with a clear map. Our experiment starts with an unbiased coin toss, which acts as the ultimate arbiter of our fate.

With a probability of $P(H) = \frac{1}{2}$ for Heads and $P(T) = \frac{1}{2}$ for Tails, the coin dictates which path we take. This is the foundation of our probability tree. We are calculating the weighted sum of two distinct possibilities.

### Phase 2: The Dice Path

If the coin lands on Heads, we enter the world of dice. We roll two unbiased dice and look for a sum of 7 or 8. The total number of outcomes is $6 	imes 6 = 36$.

To find the favorable outcomes for a sum of 7, we list them systematically: $\{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}$. This yields 6 ways.

For a sum of 8, we have $\{(2,6), (3,5), (4,4), (5,3), (6,2)\}$, which yields 5 ways. Combined, we have $6 + 5 = 11$ favorable outcomes. Thus, the conditional probability of our target event, given a Head, is:

$$P(A|H) = \frac{11}{36}$$

### Phase 3: The Card Path

Now, let us pivot to the Tails path. Here, we pick one card from a deck numbered 1 through 9. The total outcomes are simply 9.

Our target event is picking a 7 or an 8. There are only 2 favorable cards: 7 and 8. Therefore, the conditional probability of our target event, given a Tail, is:

$$P(A|T) = \frac{2}{9}$$

### Phase 4: The Synthesis

We have our two branches. Now, we use the Law of Total Probability to bring them together:

$$P(A) = P(H) \cdot P(A|H) + P(T) \cdot P(A|T)$$

Substituting our values, we get:

$$P(A) = \left(\frac{1}{2} \cdot \frac{11}{36}ight) + \left(\frac{1}{2} \cdot \frac{2}{9}ight)$$

This simplifies to:

$$P(A) = \frac{11}{72} + \frac{2}{18}$$

To add these, we find a common denominator of 72. Multiplying the second fraction by $\frac{4}{4}$, we get $\frac{8}{72}$. Finally:

$$\frac{11}{72} + \frac{8}{72} = \frac{19}{72}$$

The elegance of this result lies in the structure of the problem. By breaking it down into mutually exclusive paths, we turned a complex scenario into a simple, logical addition. You have successfully navigated the crossroads of chance, arriving at the final probability of **$\frac{19}{72}$**.

Select Answer:

Visualized Solution

Visualizing the Experiment

The Two Paths

Defining the Target Event

Case 1: Rolling the Dice (Sum 7)

Case 1: Rolling the Dice (Sum 8)

Conditional Probability for Dice

Case 2: Picking a Card

Conditional Probability for Cards

Applying Total Probability Theorem

Substitution of Values

Final Calculation

Summary and Key Takeaway

The Sigma Insight: Total Probability Theorem

The Crossroads of Chance

Phase 1

Phase 2

Phase 3

Phase 4

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