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Statistics and Probability: Mathematics | JEE Main

Q: Why does give only even powers of ?

By binomial expansion, odd powers of cancel out and even powers are doubled, so dividing by 2 leaves exactly the terms with even powers of .

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 :

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\frac{881}{2592}

\frac{27}{288}

\frac{440}{2592}

\frac{521}{2592}

Step-by-step solutionView

Correct answer

The probability that the product of the five rolls is strictly positive is $\frac{521}{2592}$ .

Option analysis

Why each option works or fails

A ·

\frac{881}{2592}

The student miscalculated the multinomial combinations or counted cases containing 0 as part of the total positive outcomes. Ensure 0 is strictly excluded from all five rolls, and carefully evaluate $\binom{5}{k} 2^k 3^{5-k}$ for each even $k \in \{0, 2, 4\}$ .

B ·

\frac{27}{288}

The student omitted the zero outcome or misclassified negative and positive counts, reducing the problem to an incorrect sample space. Recognize that there are 6 distinct faces $\{-2, -1, 0, 1, 2, 3\}$ , so the total sample space size for 5 throws is $6^5 = 7776$ .

C ·

\frac{440}{2592}

The student forgot to include the case with zero negative numbers (all 5 rolls positive) or made an arithmetic error summing the valid cases. Include all even numbers of negative outcomes: exactly 0, 2, or 4 negative numbers, with all remaining rolls being strictly positive.

D ·

\frac{521}{2592}

None. The student correctly excluded 0 and summed the cases with 0, 2, and 4 negative outcomes. The total favorable outcomes are $3^5 + \binom{5}{2} 2^2 3^3 + \binom{5}{4} 2^4 3^1 = 243 + 1080 + 240 = 1563$ , giving a probability of $\frac{1563}{6^5} = \frac{521}{2592}$ .

Reviewed route

Solution

StepWorking

01given

A die with faces $\{-2, -1, 0, 1, 2, 3\}$ is rolled 5 times. Each face has probability $1/6$ .

02goal

Find the probability that the product of the 5 outcomes is strictly positive ( $>0$ ).

03approach

Categorize the outcomes into three sets: Zero $\{0\}$ (1 outcome), Negative $\{-2, -1\}$ (2 outcomes), and Positive $\{1, 2, 3\}$ (3 outcomes). The product is positive if and only if zero never appears. Also, an even number of negative numbers must appear (0, 2, or 4 negatives). The remaining rolls must all be positive.

04execute

Case 1: 0 negative, 5 positive. Number of outcomes: $\binom{5}{0} \times 2^0 \times 3^5 = 1 \times 1 \times 243 = 243$ . Case 2: 2 negative, 3 positive. Number of outcomes: $\binom{5}{2} \times 2^2 \times 3^3 = 10 \times 4 \times 27 = 1080$ . Case 3: 4 negative, 1 positive. Number of outcomes: $\binom{5}{4} \times 2^4 \times 3^1 = 5 \times 16 \times 3 = 240$ . Sum of favorable outcomes: $243 + 1080 + 240 = 1563$ .

05execute

Total outcomes $= 6^5 = 7776$ . Required probability $= \frac{1563}{7776} = \frac{1563 / 3}{7776 / 3} = \frac{521}{2592}$ .

✓verify

521 is prime, so 521/2592 is fully simplified. Check total possibilities without zeros: $(2+3)^5 = 5^5 = 3125$ . Positive products (1563) + negative products ( $(2+3)^5 - (3-2)^5)/2 = (3125-1)/2 = 1562$ . Total $= 1563 + 1562 = 3125$ , which matches $5^5$ exactly.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Can 0 be considered positive?

No, 0 is neither positive nor negative. Any appearance of 0 makes the product 0, so 0 must never be rolled.

Why does

\frac{(P+N)^5 + (P-N)^5}{2}

give only even powers of

N

By binomial expansion, odd powers of $N$ cancel out and even powers are doubled, so dividing by 2 leaves exactly the terms with even powers of $N$ .

Answer

The probability that the product of the five rolls is strictly positive is $\frac{521}{2592}$ .

Why each option works or fails

A: $\frac{881}{2592}$ - The student miscalculated the multinomial combinations or counted cases containing 0 as part of the total positive outcomes. Ensure 0 is strictly excluded from all five rolls, and carefully evaluate $\binom{5}{k} 2^k 3^{5-k}$ for each even $k \in \{0, 2, 4\}$ .
B: $\frac{27}{288}$ - The student omitted the zero outcome or misclassified negative and positive counts, reducing the problem to an incorrect sample space. Recognize that there are 6 distinct faces $\{-2, -1, 0, 1, 2, 3\}$ , so the total sample space size for 5 throws is $6^5 = 7776$ .
C: $\frac{440}{2592}$ - The student forgot to include the case with zero negative numbers (all 5 rolls positive) or made an arithmetic error summing the valid cases. Include all even numbers of negative outcomes: exactly 0, 2, or 4 negative numbers, with all remaining rolls being strictly positive.
D · correct: $\frac{521}{2592}$ - None. The student correctly excluded 0 and summed the cases with 0, 2, and 4 negative outcomes. The total favorable outcomes are $3^5 + \binom{5}{2} 2^2 3^3 + \binom{5}{4} 2^4 3^1 = 243 + 1080 + 240 = 1563$ , giving a probability of $\frac{1563}{6^5} = \frac{521}{2592}$ .

Step-by-step solution

given: A die with faces $\{-2, -1, 0, 1, 2, 3\}$ is rolled 5 times. Each face has probability $1/6$ .
goal: Find the probability that the product of the 5 outcomes is strictly positive ( $>0$ ).
approach: Categorize the outcomes into three sets: Zero $\{0\}$ (1 outcome), Negative $\{-2, -1\}$ (2 outcomes), and Positive $\{1, 2, 3\}$ (3 outcomes). The product is positive if and only if zero never appears. Also, an even number of negative numbers must appear (0, 2, or 4 negatives). The remaining rolls must all be positive.
execute: Case 1: 0 negative, 5 positive. Number of outcomes: $\binom{5}{0} \times 2^0 \times 3^5 = 1 \times 1 \times 243 = 243$ . Case 2: 2 negative, 3 positive. Number of outcomes: $\binom{5}{2} \times 2^2 \times 3^3 = 10 \times 4 \times 27 = 1080$ . Case 3: 4 negative, 1 positive. Number of outcomes: $\binom{5}{4} \times 2^4 \times 3^1 = 5 \times 16 \times 3 = 240$ . Sum of favorable outcomes: $243 + 1080 + 240 = 1563$ .
execute: Total outcomes $= 6^5 = 7776$ . Required probability $= \frac{1563}{7776} = \frac{1563 / 3}{7776 / 3} = \frac{521}{2592}$ .
verify: 521 is prime, so 521/2592 is fully simplified. Check total possibilities without zeros: $(2+3)^5 = 5^5 = 3125$ . Positive products (1563) + negative products ( $(2+3)^5 - (3-2)^5)/2 = (3125-1)/2 = 1562$ . Total $= 1563 + 1562 = 3125$ , which matches $5^5$ exactly.

Shortcut:

When to use it: When counting even powers of negative outcomes over independent trials.

given: 5 throws of a die with 3 positive numbers, 2 negative numbers, and 1 zero.

approach: Let $P$ be a positive outcome (weight 3) and $N$ be a negative outcome (weight 2). The sum of terms with an even number of $N$ 's in $(P+N)^5$ is given by $\frac{(P+N)^5 + (P-N)^5}{2}$ .

execute: Substitute $P = 3$ and $N = 2$ : Favorable outcomes $= \frac{(3 + 2)^5 + (3 - 2)^5}{2} = \frac{5^5 + 1^5}{2} = \frac{3125 + 1}{2} = \frac{3126}{2} = 1563$ . Probability $= \frac{1563}{6^5} = \frac{521}{2592}$ .

verify: $1563/7776 = 521/2592$ . This matches the case-by-case decomposition in 10 seconds.

Statistics and Probability: Mathematics | JEE Main

What feels right?

The probability that the product of the five rolls is strictly positive is $\frac{521}{2592}$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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