Geometric Distribution Math Example 2

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Example 2

hard
For a geometric distribution with p=0.4p=0.4: (a) find P(X≤3)P(X \leq 3), (b) find the expected number of trials until first success.

Solution

  1. 1
    (a) P(X≤3)=P(X=1)+P(X=2)+P(X=3)P(X \leq 3) = P(X=1) + P(X=2) + P(X=3)
  2. 2
    P(X=1)=0.4P(X=1) = 0.4; P(X=2)=(0.6)(0.4)=0.24P(X=2) = (0.6)(0.4) = 0.24; P(X=3)=(0.6)2(0.4)=0.144P(X=3) = (0.6)^2(0.4) = 0.144
  3. 3
    P(X≤3)=0.4+0.24+0.144=0.784P(X \leq 3) = 0.4 + 0.24 + 0.144 = 0.784
  4. 4
    (b) Expected value: E(X)=1p=10.4=2.5E(X) = \frac{1}{p} = \frac{1}{0.4} = 2.5 trials on average

Answer

P(X≤3)=0.784P(X \leq 3) = 0.784; E(X)=10.4=2.5E(X) = \frac{1}{0.4} = 2.5 trials.
The geometric distribution has expected value E(X)=1/pE(X) = 1/p — intuitive since higher success probability means fewer expected trials. P(X≤k)=1−(1−p)kP(X ≤ k) = 1 - (1-p)^k (complement: probability of NOT getting all failures in k trials).

About Geometric Distribution

The probability distribution for the number of independent Bernoulli trials needed to get the first success, where each trial has success probability pp.

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