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Question
Solve the following problem :
The p.m.f. of a r.v.X is given by
`P(X = x) = {(((5),(x)) 1/2^5", ", x = 0", "1", "2", "3", "4", "5.),(0,"otherwise"):}`
Show that P(X ≤ 2) = P(X ≤ 3).
Solution
P(X ≤ 2) = P(X = 0 or X = 1 or X = 2)
= P(X = 0) + P(X = 1) + P(X = 2)
= `(""^5"C"_0)/(2^5) + (""^5"C"_1)/(2^5) + (""^5"C"_2)/(2^5)`
= `(""^5"C"_0 + ""^5"C"_1 + ""^5"C"_2)/(2^5)`
= `(1 + 5 + 10)/(2^5)`
= `(16)/(32)`
= `(1)/(2)` ...(i)
P(X ≥ 3) = P (X = 3 or X = 4 or X = 5)
= P(X = 3) + P(X = 4) + P(X = 5)
= `(""^5"C"_3)/(2^5) + (""^5"C"_4)/(2^5) + (""^5"C"_5)/(2^5)`
= `(10 + 5 + 1)/(2)`
= `(16)/(32)`
= `(1)/(2)` ...(ii)
From (i) and (ii), we get
P(X ≤ 2) = P(X ≥ 3).
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