Solution :
Here, n = 5 and r \(\ge\) 1
\(\therefore\) p(X = r) = \(^nC_r\) \(p^{n-r}\) \(q^r\)
P(X \(\ge\) 1) = 1 – P(X = 0)
= 1 – \(^5C_0 . p^5 . q^0\) \(\ge\) \(31\over 32\) [Given]
\(\implies\) \(p^5\) \(\le\) 1 – \(31\over 32\) = \(1\over 32\)
\(\therefore\) p \(\le\) \(1\over 2\) and p \(\ge\) 0
\(\implies\) p \(\in\) [0, 1/2]
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