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A physics-major table lists P(x=1)=0.35, P(x=2)=0.20, P(x=3)=0.15, P(x=5)=0.10, P(x=6)=0.05, with P(x=4) left blank. What value of P(x=4) makes this a valid discrete PDF?
A0.15
B0.35
C0.05
D0.20
Answer & Solution
Correct answer: A. 0.15
1. A valid discrete PDF must have all of its probabilities sum to exactly one.
2. Add the five known values: 0.35 + 0.20 + 0.15 + 0.10 + 0.05 = 0.85.
3. The missing P(x=4) must make the total reach one, so P(x=4) = 1 - 0.85 = 0.15.
4. The other options either repeat an existing entry or fail to bring the column total to one.
_Source: OpenStax Introductory Statistics (CC BY 4.0), Ch 4 "Discrete Random Variables", section 4.1 Probability Distribution Function (PDF) for a Discrete Random Variable_
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