Probability Theory JEE 2024: Integer-Product Solution

What is the answer to the probability theory JEE 2024 integer-product question?

The probability theory JEE 2024 integer-product question has numerical answer 8; the supplied solution identifies option 4. The probability must agree with the outcome count generated by its own product threshold.

Two positive integers total 24. Their product must reach at least a fraction of its maximum, and the probability of doing so must equal that same fraction in lowest terms. “Not less than” includes equality. Find the denominator minus the numerator:

k = \frac{m}{n}, gcd (m, n) = 1; find n - m .

Source note: The supplied bank record is tagged JEE Main 2024, but its date field says 2023-01-24. This does not independently establish the paper date or shift. The missing option list cannot be reconstructed from the supplied solution.

How do we fix the sample space and maximum product?

The supplied solution treats 23 choices as equally likely, giving a maximum product of 144. Do not switch to 12 equally weighted unordered pairs: that would change the probability model.

Write the two integers as:

x, 24 - x, x \in {1, 2, \dots, 23} .

The endpoints are excluded because both integers must be positive:

x = 0 ⟹ (0, 24), x = 24 ⟹ (24, 0) .

Reversed pairs are separate outcomes under this model. The central pair occurs only once, so unordered pairs would not all have equal weight.

Expand the product and complete the square:

f (x) & = x (24 - x) & = 24 x - x^{2} & = 144 - (x^{2} - 24 x + 144) & = 144 - (x - 12)^{2} .

A square cannot be negative. The greatest product is therefore:

f_{max} = 144, x = 12, (x, 24 - x) = (12, 12) .

How must the probability agree with its own threshold?

The proposed fraction is valid only if counting favourable outcomes produces that same fraction. Changing the proposed probability changes the threshold, so each proposed count must be checked again.

Let:

k = \frac{m}{n}, r = number of favourable values of x .

With 23 equally likely outcomes:

P (f (x) \geq 144 k) = \frac{r}{23} .

The question requires:

P (f (x) \geq 144 k) = k ⟹ k = \frac{r}{23} .

Substituting back, exactly the proposed number of integers must satisfy the inequality:

# {x \in {1, \dots, 23} : x (24 - x) \geq \frac{144 r}{23}} = r .

Propose a count, calculate its threshold, then count again. Use actual integers, not continuous interval lengths.

How does a favourable count of 15 give the answer 8?

A count of 15 works because its threshold admits exactly 15 integers. The endpoint products settle the check: the first included products are 95, while the next excluded products are 80.

Test the supplied solution’s candidate:

r = 15, k = \frac{15}{23} .

Its product threshold is:

144 \times \frac{15}{23} = \frac{2160}{23} \approx 93.913 .

Since the product is an integer:

f (x) \geq \frac{2160}{23} ⟺ f (x) \geq 94 .

This is a ceiling operation, not ordinary rounding. Use the completed square:

144 - (x - 12)^{2} \geq 94 ⟹ (x - 12)^{2} \leq 50 .

The difference from 12 is an integer, and: $7^{2} = 49 \leq 50 < 64 = 8^{2} .$

Therefore:

| x - 12 | \leq 7 ⟹ 5 \leq x \leq 19 .

Check both boundaries explicitly:

f (5) = f (19) = 5 \times 19 = 95,

f (4) = f (20) = 4 \times 20 = 80 .

Count inclusively:

19 - 5 + 1 = 15, P = \frac{15}{23} .

No other count can work. As the proposed count increases, the threshold rises, so the actual favourable count cannot increase. Below 15, the actual count is at least 15 and exceeds the proposal; above 15, it is at most 15 and falls below the proposal.

The fraction is already in lowest terms:

gcd (15, 23) = 1, m = 15, n = 23,

n - m = 23 - 15 = 8 .

Numerical answer: 8. Option number in the supplied solution: 4.

Why does the result 9 fail?

The result 9 comes from proposing 14 favourable outcomes without checking whether the resulting threshold gives 14. It does not: the successful integers remain the same 15 values found above.

The incorrect candidate would be:

r = 14, k = \frac{14}{23}, n - m = 23 - 14 = 9 .

Its threshold is:

144 \times \frac{14}{23} = \frac{2016}{23} \approx 87.652 .

The boundary products are still:

f (4) = f (20) = 80 < 87.652,

f (5) = f (19) = 95 > 87.652 .

Thus:

x \in {5, 6, \dots, 19}, P = \frac{15}{23} \neq \frac{14}{23} .

The mistake is selecting a neighbouring candidate fraction without recomputing the discrete event count. Lowering a threshold does not necessarily admit another integer.

Use this check for related problems:

Propose the fraction.
Calculate the product threshold.
Check the boundary integers.
Count inclusively.
Compare the recovered fraction with the proposal.

Which three related probability questions should I solve next?

Use these original related practice questions, not authenticated JEE PYQs, to practise fixed-threshold counting and then the self-consistency test. Attempt each count before reading its worked answer.

Question 1: Choose an integer uniformly from: $x \in {1, \dots, 23} .$

Find: $P (x (24 - x) \geq 80) .$

Worked answer:

144 - (x - 12)^{2} \geq 80 ⟹ (x - 12)^{2} \leq 64 ⟹ 4 \leq x \leq 20 .

P = \frac{20 - 4 + 1}{23} = \frac{17}{23} .

Question 2: Under the same sampling rule, find: $P (x (24 - x) \geq 96) .$

Worked answer:

(x - 12)^{2} \leq 48 ⟹ | x - 12 | \leq 6 ⟹ 6 \leq x \leq 18 .

P = \frac{18 - 6 + 1}{23} = \frac{13}{23} .

Compare the thresholds with the main question:

At 80, both outer boundary products are included.
At approximately 93.913, those two outcomes are excluded.
At 96, the two products equal to 95 are also excluded.

A small threshold change can add or remove two symmetric outcomes. Probability changes in discrete jumps.

Question 3: Choose an integer uniformly from the set below, forming two positive integers with sum 12:

x \in {1, \dots, 11}, (x, 12 - x) .

Find the requested difference if:

P (x (12 - x) \geq 36 k) = k = \frac{m}{n}, gcd (m, n) = 1 .

$Find n - m .$ Worked answer: Verify the candidate:

k = \frac{7}{11}, 36 k = \frac{252}{11} \approx 22.909 .

The product has the form:

x (12 - x) = 36 - (x - 6)^{2} .

The qualifying integers and adjacent boundary products are:

3 \leq x \leq 9, 3 \times 9 = 27, 2 \times 10 = 20 .

There are seven successes, so:

P = \frac{9 - 3 + 1}{11} = \frac{7}{11}, n - m = 11 - 7 = 4 .

The same increasing-threshold argument proves uniqueness. On your next attempt, write the two boundary products before accepting any proposed fraction.

Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

For a worked example of the same idea, see Chemical Equilibrium JEE 2016: Why Option C Is Incorrect.

Frequently asked questions

What is the answer to the JEE 2024 integer-product probability question?

The numerical answer is 8. The threshold corresponding to 15/23 admits exactly 15 of the 23 equally likely outcomes, so the required fraction is m/n = 15/23. It is already in lowest terms, giving n − m = 23 − 15 = 8.

Why are there 23 outcomes instead of 12 unordered pairs?

The supplied solution chooses x uniformly from 1 to 23 and forms the pair (x, 24 − x). Reversed pairs are separate outcomes, while (12, 12) occurs only once. Treating the 12 unordered pairs as equally likely would change the probability model.

How do you find the maximum product of two positive integers with sum 24?

Write the integers as x and 24 − x, so their product is 144 − (x − 12)². The square is non-negative, making the maximum product 144 at x = 12. Both integers are therefore 12.

Why is the answer 8 and not 9?

The answer 9 would require the fraction 14/23, whose product threshold is 2016/23, approximately 87.652. That threshold still admits every integer from 5 to 19, giving 15 favourable outcomes rather than 14. The recovered probability is 15/23, so 14/23 fails the required self-consistency check.