Each attempt to fill an order succeeds independently with chance p, so the number of attempts X has P(X = x) = (1 − p)^(x − 1) × p for x = 1, 2, …. Across 5 orders the attempts needed were 2, 5, 1, 4 and 3. What is the maximum likelihood estimate of p? Give a fraction.

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Each attempt to fill an order succeeds independently with chance $p$ , so the number of attempts $X$ has $P (X = x) = (1 - p)^{x - 1} \times p$ for $x = 1, 2, \dots$ . Across $5$ orders the attempts needed were $2$ , $5$ , $1$ , $4$ and $3$ . What is the maximum likelihood estimate of $p$ ? Give a fraction.

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Hint

The log likelihood counts $n$ successes and $\sum (x_{i} - 1)$ failures.

Solution

There are $5$ successes, and $2 + 5 + 1 + 4 + 3 = 15$ attempts, so $15 - 5 = 10$ failures.
The slope of $5 log p + 10 log (1 - p)$ is $\frac{5}{p} - \frac{10}{1 - p}$ , which is zero when $5 \times (1 - p) = 10 \times p$ .
That gives $p^= \frac{5}{15} = \frac{1}{3}$ .

Answer: $\frac{1}{3}$

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