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Q.A random variable X has P(X=k) = (10k)210\dfrac{\binom{10}{k}}{2^{10}} for k=0,1,…,10. Then P(X ≤ 5) equals

a
12\dfrac{1}{2}
b
12+(105)211\dfrac{1}{2} + \dfrac{\binom{10}{5}}{2^{11}}
c
12−(105)211\dfrac{1}{2} - \dfrac{\binom{10}{5}}{2^{11}}
d
(105)210\dfrac{\binom{10}{5}}{2^{10}}

Correct Answer: Option B

The correct solution involves applying the fundamental concept to derive the final value step by step...

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