If an unbiased die, marked with -2,-1,0,1,2,3 on its faces, is thrown five times, then the probability that the product of the outcomes is positive, is :

Probability PYQ — JEE Main 2023 · 30 Jan · Shift 1 · Q32 with Solution

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Correct answer: B

Classify the outcomes of one throw

The die has faces marked: $-2,-1,0,1,2,3$

So for one throw:

Negative outcomes: $-2,-1$ outcomes
Zero outcome: $0$ 1 outcome
Positive outcomes: $1,2,3$ 3 outcomes

Thus,

\quad P(\text{zero})=\frac{1}{6}, \quad P(\text{positive})=\frac{3}{6}=\frac12.$$ 2. **Condition for product to be positive** The product of the five outcomes is positive if and only if: - no throw is $0$, and - the number of negative outcomes is even. Since there are $5$ throws, the possible even numbers of negatives are: $$0,2,4.$$ 3. **Count favorable cases** Total number of outcomes in $5$ throws: $$6^5=7776.$$ We count favorable sequences according to the number of negative outcomes. --- ### Case 1: 0 negatives Then all 5 outcomes must be positive. Each positive can be chosen in $3$ ways, so number of sequences: $$3^5=243.$$ --- ### Case 2: 2 negatives Choose the 2 positions of negatives: $$\binom52=10.$$ For each negative position, there are $2$ choices ($-2,-1$), so: $$2^2=4.$$ For each of the remaining 3 positions, outcome must be positive, so: $$3^3=27.$$ Hence number of sequences: $$\binom52\cdot 2^2\cdot 3^3=10\cdot 4\cdot 27=1080.$$ --- ### Case 3: 4 negatives Choose the 4 positions of negatives: $$\binom54=5.$$ Negative values can be chosen in: $$2^4=16$$ ways. The remaining 1 position must be positive, chosen in: $$3$$ ways. Hence number of sequences: $$\binom54\cdot 2^4\cdot 3=5\cdot 16\cdot 3=240.$$ --- 4. **Total favorable outcomes** $$243+1080+240=1563.$$ Therefore, $$P(\text{product positive})=\frac{1563}{7776}.$$ Simplify: $$\frac{1563}{7776}=\frac{521}{2592}.$$ 5. **Compare with options** The probability is $$\boxed{\frac{521}{2592}}$$ which matches **Option B**. 6. **Comparison with stored correct answer** Stored correct answer: **B** Our derived answer is also **B**, so they agree.

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View written solutionFree

Correct answer: B

Classify the outcomes of one throw

The die has faces marked: $-2,-1,0,1,2,3$

So for one throw:

Negative outcomes: $-2,-1$ outcomes
Zero outcome: $0$ 1 outcome
Positive outcomes: $1,2,3$ 3 outcomes

Thus,

\quad P(\text{zero})=\frac{1}{6}, \quad P(\text{positive})=\frac{3}{6}=\frac12.$$ 2. **Condition for product to be positive** The product of the five outcomes is positive if and only if: - no throw is $0$, and - the number of negative outcomes is even. Since there are $5$ throws, the possible even numbers of negatives are: $$0,2,4.$$ 3. **Count favorable cases** Total number of outcomes in $5$ throws: $$6^5=7776.$$ We count favorable sequences according to the number of negative outcomes. --- ### Case 1: 0 negatives Then all 5 outcomes must be positive. Each positive can be chosen in $3$ ways, so number of sequences: $$3^5=243.$$ --- ### Case 2: 2 negatives Choose the 2 positions of negatives: $$\binom52=10.$$ For each negative position, there are $2$ choices ($-2,-1$), so: $$2^2=4.$$ For each of the remaining 3 positions, outcome must be positive, so: $$3^3=27.$$ Hence number of sequences: $$\binom52\cdot 2^2\cdot 3^3=10\cdot 4\cdot 27=1080.$$ --- ### Case 3: 4 negatives Choose the 4 positions of negatives: $$\binom54=5.$$ Negative values can be chosen in: $$2^4=16$$ ways. The remaining 1 position must be positive, chosen in: $$3$$ ways. Hence number of sequences: $$\binom54\cdot 2^4\cdot 3=5\cdot 16\cdot 3=240.$$ --- 4. **Total favorable outcomes** $$243+1080+240=1563.$$ Therefore, $$P(\text{product positive})=\frac{1563}{7776}.$$ Simplify: $$\frac{1563}{7776}=\frac{521}{2592}.$$ 5. **Compare with options** The probability is $$\boxed{\frac{521}{2592}}$$ which matches **Option B**. 6. **Comparison with stored correct answer** Stored correct answer: **B** Our derived answer is also **B**, so they agree.

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