Count All Sequences with an Exact Success Count

Supplied materials

• Paper or a browser scratch area; a calculator is optional unless rounding is requested.
• Supplied teaching case: A fair coin is tossed independently 3 times. Find the probability of exactly one head. Candidate answers: (A) 1/2 (B) 1/8 (C) 3/8 (D) 7/8.
• Independent check inputs: A fair coin is tossed independently 4 times. Find the probability of exactly one head. Candidate answers: (A) 1/16 (B) 1/2 (C) 1/4 (D) 15/16.

Learn the idea

Counting methods distinguish ordered roles from unordered groups. A probability model assigns chances to possible outcomes, and expected value or binomial counting combines those probabilities while preserving what a single trial can actually produce. Repeated independent two-outcome trials create ordered sequences. Exactly one success can occupy any one trial position; the resulting complete sequence probabilities can be added.

Here is the method on the supplied teaching case. The single head can occupy any of 3 positions. Each resulting complete sequence has probability (1/2)^3; adding these disjoint sequences gives 3/8. Exactly one excludes all other head counts.

Exactly one excludes every other success count. Independence and equal trial probability are explicit here; changing either assumption changes the sequence probabilities and invalidates the simple equal-weight count.

1. Write or sketch the supplied teaching case, labeling its quantities and restrictions.
2. Apply the method shown in the worked solution, keeping each intermediate equality or justified inference.
3. Solve the independent check with its changed inputs before revealing feedback.
4. Complete the three practice cases; explain which mathematical rule rules out one distractor in each case.

Check your reasoning

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Take it further

Changed-input task: A fair coin is tossed independently 10 times. Find the probability of exactly one head. Candidate answers: (A) 1/1024 (B) 1/2 (C) 1023/1024 (D) 5/512. Show the mathematical reason, not only the selected answer. Expected result: 5/512. The single head can occupy any of 10 positions. Each resulting complete sequence has probability (1/2)^10; adding these disjoint sequences gives 5/512. Exactly one excludes all other head counts.

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