Geometric Distribution Math Example 3

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Example 3

easy
A fair coin is flipped until heads appears. What is P(first heads on flip 4)P(\text{first heads on flip 4}), and what is the expected number of flips?

Solution

  1. 1
    p=0.5p = 0.5; P(X=4)=(0.5)4−1×0.5=(0.5)4=116=0.0625P(X=4) = (0.5)^{4-1} \times 0.5 = (0.5)^4 = \frac{1}{16} = 0.0625
  2. 2
    Expected flips: E(X)=1p=10.5=2E(X) = \frac{1}{p} = \frac{1}{0.5} = 2 flips on average

Answer

P(X=4)=116=0.0625P(X=4) = \frac{1}{16} = 0.0625; expected flips = 2.
For a fair coin, we expect to wait 2 flips on average for the first heads. The probability decreases exponentially for later first-successes: P(X=1)=0.5, P(X=2)=0.25, P(X=3)=0.125, P(X=4)=0.0625.

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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More Geometric Distribution Examples

Example 1 medium

A basketball player makes free throws with probability [formula]. Find the probability they make the

Example 2 hard

For a geometric distribution with [formula]: (a) find [formula], (b) find the expected number of tri

Example 4 hard

A quality inspector samples items until finding the first defective. Defect probability is [formula]

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