Sum of probabilities =1: 302a+1+8a−1+4a+1+b=1⇒b=3029−14a.
σ2+μ2=E(X2)=2.
E(X2)=308a−1+304(4a+1)+309(29−14a)=30−102a+264=2.
−102a+264=60⇒a=2.
b=3029−28=301.
ba=1/302=60.
JEE Main 2026 — Mathematics Probability & Statistics
A random variable X takes values 0,1,2,3 with probabilities 302a+1,308a−1,304a+1,b respectively, where a,b∈R. Let μ and σ respectively be the mean and standard deviation of X such that σ2+μ2=2. Then ba is equal to :
Held on 21 Jan 2026 · Verified 6 Jul 2026.
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