MCQEasyJEE 2026Probability Basics

JEE Mathematics 2026 Question with Solution

Two distinct numbers aa and bb are selected at random from 1,2,3,,501, 2, 3, \ldots, 50. The probability that their product abab is divisible by 33 is

  • A

    825\dfrac{8}{25}

  • B

    5611225\dfrac{561}{1225}

  • C

    6641225\dfrac{664}{1225}

  • D

    2721225\dfrac{272}{1225}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Two distinct numbers aa and bb are chosen from {1,2,3,,50}\{1,2,3,\ldots,50\}.

Find: The probability that abab is divisible by 33.

The total number of ways to choose two distinct numbers is

(502)=50×492=1225\binom{50}{2} = \frac{50 \times 49}{2} = 1225

Numbers divisible by 33 in {1,2,3,,50}\{1,2,3,\ldots,50\} are 3,6,9,,483,6,9,\ldots,48, so their count is

503=16\left\lfloor \frac{50}{3} \right\rfloor = 16

Hence, the count of numbers not divisible by 33 is

5016=3450 - 16 = 34

Use the complementary event: abab is not divisible by 33 only when neither aa nor bb is divisible by 33.

So the number of such selections is

(342)=34×332=561\binom{34}{2} = \frac{34 \times 33}{2} = 561

Therefore, the number of favorable selections is

1225561=6641225 - 561 = 664

So the required probability is

6641225\frac{664}{1225}

The solution lists option B and final answer 5611225\dfrac{561}{1225}, but the working clearly gives 6641225\dfrac{664}{1225}. Therefore, the correct option from the given choices is C.

Complement Counting Shortcut

Given: Two distinct numbers are selected from 11 to 5050.

Find: Probability that the product is divisible by 33.

Instead of directly counting cases where at least one number is divisible by 33, count the opposite case first.

There are 1616 multiples of 33 up to 5050, so there are 3434 numbers not divisible by 33. The product fails to be divisible by 33 only if both chosen numbers come from these 3434 numbers.

Thus,

P(not divisible by 3)=(342)(502)=5611225P(\text{not divisible by } 3) = \frac{\binom{34}{2}}{\binom{50}{2}} = \frac{561}{1225}

Hence,

P(divisible by 3)=15611225=6641225P(\text{divisible by } 3) = 1 - \frac{561}{1225} = \frac{664}{1225}

Therefore, the correct option is C.

Common mistakes

  • Counting 5611225\dfrac{561}{1225} as the required probability. This is the probability that neither selected number is divisible by 33, so the product is not divisible by 33. Use the complement to get the required probability.

  • Using ordered pairs instead of unordered selections. The question selects two distinct numbers, so the total outcomes are (502)\binom{50}{2}, not 50×4950 \times 49.

  • Miscounting the multiples of 33 up to 5050. The sequence ends at 4848, so the correct count is 1616, obtained from 503\left\lfloor \frac{50}{3} \right\rfloor.

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