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 1 \over n . If the die is thrown thrice, then the probability, that the sum of the numbers obtained is 48, is :

Probability PYQ — JEE Main 2022 · 25 Jun · Shift 2 · Q34 with Solution

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Correct answer: D

Find the probability distribution of the die

The faces are marked: $2,4,8,16,32,32$

Given: probability of getting a face with mark $n$ is proportional to $\frac1n$ . So for a face marked $n$ , $P(n)=k\cdot \frac1n$ for some constant $k$ .

Since there are two faces marked $32$ , each such face has probability $k\cdot \frac1{32}$ . Thus total probability is $k\left(\frac12+\frac14+\frac18+\frac1{16}+\frac1{32}+\frac1{32}\right)=1$

Now,

=\frac{16+8+4+2+1+1}{32}=\frac{32}{32}=1$$ Hence, $$k=1$$ Therefore, $$P(2)=\frac12,\quad P(4)=\frac14,\quad P(8)=\frac18,\quad P(16)=\frac1{16}$$ For the two faces marked $32$, each has probability $\frac1{32}$, so total $$P(32)=\frac1{32}+\frac1{32}=\frac1{16}$$ So the effective distribution is: $$P(2)=\frac12,\; P(4)=\frac14,\; P(8)=\frac18,\; P(16)=\frac1{16},\; P(32)=\frac1{16}$$ --- 2. **Find all ways to get sum $48$ in three throws** We need triples $(a,b,c)$ from $\{2,4,8,16,32\}$ such that $$a+b+c=48$$ Since all numbers are powers of $2$, check possible combinations: - If one throw is $32$, then the other two must sum to $16$. Possible: $$8+8=16$$ So one combination is $$(32,8,8)$$ - Also with one throw $32$, the other two could be $$2+14,\;4+12,\;16+0$$ but these are not possible. - Without $32$, largest possible sum is $$16+16+16=48$$ giving another combination: $$(16,16,16)$$ No other combinations are possible. So favorable cases are: 1. $(32,8,8)$ and its permutations 2. $(16,16,16)$ --- 3. **Compute probability of each case** ### Case 1: $(32,8,8)$ Probability of one specific order, say $(32,8,8)$, is $$P(32)P(8)P(8)=\frac1{16}\cdot \frac18\cdot \frac18=\frac1{1024}=\frac1{2^{10}}$$ Number of distinct permutations of $(32,8,8)$ is $$\frac{3!}{2!}=3$$ So total probability for this case is $$3\cdot \frac1{2^{10}}=\frac3{2^{10}}$$ ### Case 2: $(16,16,16)$ Probability is $$P(16)^3=\left(\frac1{16}\right)^3=\frac1{4096}=\frac1{2^{12}}$$ --- 4. **Add the probabilities** $$P(\text{sum }=48)=\frac3{2^{10}}+\frac1{2^{12}}$$ Convert to common denominator $2^{12}$: $$\frac3{2^{10}}=\frac{12}{2^{12}}$$ Hence, $$P=\frac{12}{2^{12}}+\frac1{2^{12}}=\frac{13}{2^{12}}$$ --- 5. **Match with the options** $$\boxed{\frac{13}{2^{12}}}$$ This is **Option D**. --- 6. **Compare with stored correct answer** Stored correct answer: **D** My derived answer is also **D**, so they agree.

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View written solutionFree

Correct answer: D

Find the probability distribution of the die

The faces are marked: $2,4,8,16,32,32$

Given: probability of getting a face with mark $n$ is proportional to $\frac1n$ . So for a face marked $n$ , $P(n)=k\cdot \frac1n$ for some constant $k$ .

Since there are two faces marked $32$ , each such face has probability $k\cdot \frac1{32}$ . Thus total probability is $k\left(\frac12+\frac14+\frac18+\frac1{16}+\frac1{32}+\frac1{32}\right)=1$

Now,

=\frac{16+8+4+2+1+1}{32}=\frac{32}{32}=1$$ Hence, $$k=1$$ Therefore, $$P(2)=\frac12,\quad P(4)=\frac14,\quad P(8)=\frac18,\quad P(16)=\frac1{16}$$ For the two faces marked $32$, each has probability $\frac1{32}$, so total $$P(32)=\frac1{32}+\frac1{32}=\frac1{16}$$ So the effective distribution is: $$P(2)=\frac12,\; P(4)=\frac14,\; P(8)=\frac18,\; P(16)=\frac1{16},\; P(32)=\frac1{16}$$ --- 2. **Find all ways to get sum $48$ in three throws** We need triples $(a,b,c)$ from $\{2,4,8,16,32\}$ such that $$a+b+c=48$$ Since all numbers are powers of $2$, check possible combinations: - If one throw is $32$, then the other two must sum to $16$. Possible: $$8+8=16$$ So one combination is $$(32,8,8)$$ - Also with one throw $32$, the other two could be $$2+14,\;4+12,\;16+0$$ but these are not possible. - Without $32$, largest possible sum is $$16+16+16=48$$ giving another combination: $$(16,16,16)$$ No other combinations are possible. So favorable cases are: 1. $(32,8,8)$ and its permutations 2. $(16,16,16)$ --- 3. **Compute probability of each case** ### Case 1: $(32,8,8)$ Probability of one specific order, say $(32,8,8)$, is $$P(32)P(8)P(8)=\frac1{16}\cdot \frac18\cdot \frac18=\frac1{1024}=\frac1{2^{10}}$$ Number of distinct permutations of $(32,8,8)$ is $$\frac{3!}{2!}=3$$ So total probability for this case is $$3\cdot \frac1{2^{10}}=\frac3{2^{10}}$$ ### Case 2: $(16,16,16)$ Probability is $$P(16)^3=\left(\frac1{16}\right)^3=\frac1{4096}=\frac1{2^{12}}$$ --- 4. **Add the probabilities** $$P(\text{sum }=48)=\frac3{2^{10}}+\frac1{2^{12}}$$ Convert to common denominator $2^{12}$: $$\frac3{2^{10}}=\frac{12}{2^{12}}$$ Hence, $$P=\frac{12}{2^{12}}+\frac1{2^{12}}=\frac{13}{2^{12}}$$ --- 5. **Match with the options** $$\boxed{\frac{13}{2^{12}}}$$ This is **Option D**. --- 6. **Compare with stored correct answer** Stored correct answer: **D** My derived answer is also **D**, so they agree.

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