A biased die is marked with numbers 2, 4, 8, 16, 32, 32 on its faces and the probability of getting

Q: Wait, why is $P(32) = \frac{1}{16}$? The problem says $P(n) = \frac{1}{n}$, so shouldn't it be $\frac{1}{32}$?

That is the classic trap! The rule $P(n) = \frac{1}{n}$ applies to each individual face. Since there are two faces marked $32$, you have two mutually exclusive events: rolling the first $32$-face OR rolling the second $32$-face. Thus, $P(32) = P(\text{face 1}) + P(\text{face 2}) = \frac{1}{32} + \frac{1}{32} = \frac{2}{32} = \frac{1}{16}$. Always count your faces before you calculate your probabilities!

Q: Why do we multiply the probability of $(32, 8, 8)$ by $3$? I thought we just multiply the probabilities of the individual numbers.

Great question. When we calculate $P(32) \times P(8) \times P(8)$, we are finding the probability of one specific sequence, like $(32, 8, 8)$. But the problem asks for the sum to be $48$, which can happen as $(32, 8, 8)$, $(8, 32, 8)$, or $(8, 8, 32)$. Since these are distinct, mutually exclusive sequences, we must add their probabilities together. Since each has the same probability, we multiply by the number of arrangements, which is $\frac{3!}{2!} = 3$.

Q: How do I know I haven't missed any combinations for the sum $48$?

Systematic listing is your best friend. Start with the largest possible value, $32$. If one die is $32$, the remaining two must sum to $16$. The only pair from our set that sums to $16$ is $(8, 8)$. If we don't use $32$, the largest is $16$. If we use $16$, the remaining two must sum to $32$. The only way is $(16, 16)$. If we try to use anything smaller, we run out of 'sum' capacity. Trust your logic, but always verify!

Question

A biased die is marked with numbers 2, 4, 8, 16, 32, 32 on its faces and the probability of getting a face with mark $n$ is $\frac{1}{n}$. If the die is thrown thrice, then the probability, that the sum of the numbers obtained is 48, is

Options:

(A) $\frac{7}{2^{11}}$

(B) $\frac{7}{2^{12}}$

(C) $\frac{3}{2^{10}}$

(D) $\frac{13}{2^{12}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The die is marked with the set $\{2, 4, 8, 16, 32, 32\}$. The probability of rolling a specific face marked $n$ is defined as $P(n) = \frac{1}{n}$.

Because the number $32$ appears on two distinct faces, we must sum their individual probabilities to find the total probability of rolling a $32$.

$$P(32) = P(	ext{face 1}) + P(	ext{face 2}) = \frac{1}{32} + \frac{1}{32} = \frac{2}{32} = \frac{1}{16}$$

The complete probability distribution for the outcomes is:
- $P(2) = \frac{1}{2}$
- $P(4) = \frac{1}{4}$
- $P(8) = \frac{1}{8}$
- $P(16) = \frac{1}{16}$
- $P(32) = \frac{1}{16}$

### The Search for 48

We perform three independent throws, $x_1, x_2, x_3$, and require the sum $x_1 + x_2 + x_3 = 48$. We identify the valid combinations of outcomes that satisfy this condition.

If we include the value $32$, the remaining two dice must sum to $16$. The only combination from our set that sums to $16$ is $(8, 8)$, yielding the set $\{32, 8, 8\}$.

If we do not include $32$, the largest available value is $16$. To reach a sum of $48$ using three values no larger than $16$, the only possibility is $\{16, 16, 16\}$.

### Calculating Probabilities

We now calculate the probability for each mutually exclusive case.

**Case 1: The outcome is $(16, 16, 16)$**
Since the throws are independent, we calculate:
$$P(16, 16, 16) = P(16) 	imes P(16) 	imes P(16) = \frac{1}{16} 	imes \frac{1}{16} 	imes \frac{1}{16} = \frac{1}{2^4} 	imes \frac{1}{2^4} 	imes \frac{1}{2^4} = \frac{1}{2^{12}}$$

**Case 2: The outcome is a permutation of $(32, 8, 8)$**
There are $\frac{3!}{2!} = 3$ distinct arrangements for this set: $(32, 8, 8)$, $(8, 32, 8)$, and $(8, 8, 32)$. The probability for this case is:
$$P(	ext{Case 2}) = 3 	imes \left[ P(32) 	imes P(8) 	imes P(8) ight]$$

Substituting the values in powers of $2$:
$$P(	ext{Case 2}) = 3 	imes \left[ \frac{1}{2^4} 	imes \frac{1}{2^3} 	imes \frac{1}{2^3} ight] = 3 	imes \frac{1}{2^{10}}$$

### Final Calculation

To sum the probabilities, we express $P(	ext{Case 2})$ with a common denominator of $2^{12}$:
$$P(	ext{Case 2}) = 3 	imes \frac{2^2}{2^{10} 	imes 2^2} = \frac{12}{2^{12}}$$

The total probability is the sum of the two cases:
$$	ext{Total Probability} = \frac{1}{2^{12}} + \frac{12}{2^{12}} = \frac{13}{2^{12}}$$

Calculating the final value, we get:
$$\mathbf{\frac{13}{4096}}$$

Animated Solution for Mathematics - Probability: A biased die is marked with numbers 2, 4, 8, 16, 32, 32 on its faces and the probability of getting a face with mark $n$ is $\frac{1}{n}$ . If the die is thrown thrice, then the probability, that the sum of the numbers obtained is 48, is

Select Answer:

Visualized Solution

Understanding the Biased Die

The Probability Rule

Calculating Individual Probabilities

Setting Up the Condition

Finding Combinations for Sum 48

Probability of Case 1

Probability of Case 2 (Setup)

Probability of Case 2 (Calculation)

Making Denominators Equal

Final Probability

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Search for 48

Calculating Probabilities

Final Calculation

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