JEE Main 2021 · 27 August · evening · Numerical answer · +4 / −0
The probability distribution of random variable X is given by :
Let p = P(1 < X < 4 | X < 3). If 5p = λK, then λ equal to ___________.
| X | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| P(X) | K | 2K | 2K | 3K | K |
Answer and solution
Answer 30
Solution
Maths sir · 4 steps
Fix K from the total probability first, then the conditional reduces to just the outcome X = 2.
- Find K. All probabilities add to 1: K + 2K + 2K + 3K + K = 9K = 1, so K = 1/9.
p = P(1 < X < 4 | X < 3) = P(1 < X < 4 and X < 3)/P(X < 3)
- Overlap and condition. 1 < X < 4 means X = 2, 3, and X < 3 means X = 1, 2. Both hold only for X = 2, so the numerator is 2K. The denominator is P(X=1) + P(X=2) = K + 2K = 3K.
- Compute p. p = 2K/3K = 2/3, so 5p = 10/3.
- So. 10/3 = λ K = λ/9, which gives λ = 10/3 × 9 = 30.
The common slip: 60. Using P(1 < X < 4) = 4K as the numerator instead of the overlap gives p = 4/3 and λ = 60. A probability above 1 is the warning sign.
The idea to keep. For a conditional probability, the numerator is always the overlap of the two events, never the first event alone. And always pin down the unknown constant using total probability = 1.
Answer from the NTA final answer key. Solution by PrepWiser.