Maths Q8 – JEE-MAIN 2025

If the probability that the random variable

X

takes the value

x

is given by

P (X = x) = k (x + 1) 3^{- x}, x = 0, 1, 2, 3, . . ., where k is a constant, then P (X \geq 3) is equal to

Clue 1 of 3: Probability Distribution Property

For any discrete probability distribution, the sum of all probabilities must be equal to 1.

🧠 Full Solution Path

Step 1: Determine the constant k✦ Active

For a valid probability distribution, the sum of all probabilities must be 1. Given $P (X = x) = k (x + 1) 3^{- x}$ , we have:

\sum_{x = 0}^{\infty} P (X = x) = 1 ⟹ \sum_{x = 0}^{\infty} k (x + 1) 3^{- x} = 1

Factor out $k$ and recognize the series $\sum_{x = 0}^{\infty} (x + 1) r^{x}$ with $r = \frac{1}{3}$ . This sum is equal to $\frac{1}{(1 - r)^{2}}$ .

k \sum_{x = 0}^{\infty} (x + 1) {(\frac{1}{3})}^{x} = k \cdot \frac{1}{{(1 - \frac{1}{3})}^{2}} = k \cdot \frac{1}{{(\frac{2}{3})}^{2}} = k \cdot \frac{1}{\frac{4}{9}} = k \cdot \frac{9}{4}

Setting this sum to 1, we find $k$ :

k \cdot \frac{9}{4} = 1 ⟹ k = \frac{4}{9}

Step 2: Calculate P(X < 3)○ Expand

To find $P (X \geq 3)$ , it's easier to use the complementary probability: $P (X \geq 3) = 1 - P (X < 3)$ . We need to calculate $P (X = 0)$ , $P (X = 1)$ , and $P (X = 2)$ .

P (X = 0) = k (0 + 1) 3^{- 0} = k = \frac{4}{9}

P (X = 1) = k (1 + 1) 3^{- 1} = 2 k \cdot \frac{1}{3} = \frac{2}{3} \cdot \frac{4}{9} = \frac{8}{27}

P (X = 2) = k (2 + 1) 3^{- 2} = 3 k \cdot \frac{1}{9} = \frac{k}{3} = \frac{1}{3} \cdot \frac{4}{9} = \frac{4}{27}

Now, sum these probabilities:

P (X < 3) = P (X = 0) + P (X = 1) + P (X = 2) = \frac{4}{9} + \frac{8}{27} + \frac{4}{27}

Find a common denominator (27):

P (X < 3) = \frac{12}{27} + \frac{8}{27} + \frac{4}{27} = \frac{12 + 8 + 4}{27} = \frac{24}{27}

Step 3: Calculate P(X >= 3)○ Expand

Finally, use the complementary probability formula:

P (X \geq 3) = 1 - P (X < 3) = 1 - \frac{24}{27} = \frac{27 - 24}{27} = \frac{3}{27} = \frac{1}{9}

💡 Teacher's Secret Hint

Ensure all fractions are simplified to their lowest terms.

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Maths Question 8 – JEE-MAIN 2025

Clue 1 of 3: Probability Distribution Property

🧠 Full Solution Path