Yes. An expected value describes a probability-weighted average, not necessarily an outcome that one trial can produce. A hypothetical packet model makes the distinction clear: each packet contains either 0 markers or 2 markers. The packet contains 0 markers with probability 1/4 and 2 markers with probability 3/4.
Calculating the expected number
Expected value sums value multiplied by probability; see the OpenStax explanation of expected value.
For the hypothetical packet:
E(markers) = 0 × 1/4 + 2 × 3/4 = 0 + 6/4 = 1.5 markers.
The result, 1.5, is valid even though a single packet cannot contain 1.5 markers. The calculation combines the two possible outcomes according to their probabilities. It does not claim that a packet with 1.5 markers is one of the permitted outcomes.
Why an expected total can also be impossible
Now consider 10 hypothetical independent packets, each following the same model. The expected total is the number of packets multiplied by the expected number per packet:
10 × 1.5 = 15 markers.
However, each packet contributes either 0 or 2 markers, so the total must be even. The complete set of possible totals is 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, or 20. Therefore, 15 is not a permitted total in this model, even though it is the correct expected total.
Average meaning versus single-outcome meaning
For these packets, 1.5 is a weighted average. It is not a promise about one packet, one group of 10 packets, or any particular trial. In this example, an illustrative actual result of 20 markers would be possible, while 15 would be impossible; neither fact invalidates the expectation.
The key question is therefore: “What values can one trial actually take?” That list describes permitted outcomes. The expected value answers a different question: “What probability-weighted average does the model produce?”
For a different question about interpreting past observations, see our history-screen reading note. The hypothetical counts above are not observed results.