Expected Profit per Widget

Expectation · Easy · Free problem

A factory produces widgets. Each widget is independently defective with probability

/50$. The company earns $\
$ profit on each non-defective widget and loses $\
$ on each defective widget (due to rework and scrap costs).

What is the expected profit per widget?

Hints

  1. This is a two-outcome expected value problem. Just identify the payoff and probability for each case.
  2. Expected profit $= (\text{profit if good}) \times P(\text{good}) + (\text{loss if defective}) \times P(\text{defective})$.
  3. Compute \times \frac{49}{50} + (-1) \times \frac{1}{50}$.

Worked Solution

How to Think About It: This is a straightforward expected value calculation. You have two outcomes with known probabilities and known payoffs -- just take the weighted average. The defect rate is low ( \%$), so the answer should be close to $\ $ but slightly less.

Quick Estimate: You make $\ $ on $98\%$ of widgets and lose $\

$ on
\%$. The loss from defectives is tiny: $0.02 \times \
= \$0.02$. So expected profit is roughly $\
- 0.02 - 0.02 \times 2 = \
.94$. Let's be precise.

Formal Solution:

Let $D$ be the event that a widget is defective. We have $P(D) = 1/50$ and $P(D^c) = 49/50$.

The profit $X$ for a single widget is: - $X = +2$ if not defective (probability $49/50$) - $X = -1$ if defective (probability

/50$)

The expected profit is:

$E[X] = 2 \cdot \frac{49}{50} + (-1) \cdot \frac{1}{50} = \frac{98}{50} - \frac{1}{50} = \frac{97}{50} = 1.94$

Answer: The expected profit per widget is $\

.94$.

Intuition

This is about as basic as expected value gets, but it illustrates an important principle: even a small defect rate has a measurable impact on expected profit. Here the \%$ defect rate knocks $\$0.06$ off the $\ $ ideal profit -- a $3\%$ haircut. In manufacturing and trading alike, the key insight is that the expected cost of rare bad outcomes is (probability) times (severity), and you should always do this multiplication explicitly rather than dismissing low-probability events as negligible.

In a quant context, this same structure appears everywhere: expected P&L from a trade is (probability of winning) times (win size) minus (probability of losing) times (loss size). The widget problem is just the simplest version of that calculation.

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