A certain club has 10 members, including Harry. One of the 10 members is to be chosen at random to...

1. Translate the problem requirements: We need to find the probability that Harry becomes either secretary OR treasurer (not president). The positions are filled sequentially: president first, then secretary from remaining 9, then treasurer from remaining 8.
2. Identify the constraint for Harry's eligibility: For Harry to be secretary or treasurer, he must NOT be chosen as president first. Determine the probability Harry avoids the president position.
3. Calculate Harry's chances in remaining positions: Given Harry wasn't chosen as president, find the probability he's selected as either secretary or treasurer from the remaining candidates.
4. Apply probability combination logic: Combine the probability that Harry avoids presidency with his probability of being chosen for one of the two desired positions.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what's happening in plain English. We have a club with 10 members, and Harry is one of them. The club needs to fill three positions:

• First, they randomly pick 1 person out of 10 to be president
• Then, they randomly pick 1 person out of the remaining 9 to be secretary
• Finally, they randomly pick 1 person out of the remaining 8 to be treasurer

The question asks: What's the probability that Harry becomes either the secretary OR the treasurer? Notice that this means Harry cannot be president - if he's president, he can't be secretary or treasurer.

2. Identify the constraint for Harry's eligibility

Here's the key insight: For Harry to have a chance at being secretary or treasurer, he must first NOT be chosen as president.

Think about it this way: When they pick the president, there are 10 people to choose from. Harry is 1 of those 10 people. So the chance that Harry is NOT picked as president is 9 out of 10, or \(\frac{9}{10}\).

If Harry gets picked as president (which happens \(\frac{1}{10}\) of the time), then he's automatically out of the running for secretary or treasurer. We only care about the scenarios where Harry avoids being president.

3. Calculate Harry's chances in remaining positions

Now let's assume Harry wasn't picked as president. This means there are 9 people left, including Harry, and they need to fill secretary and treasurer positions.

Here's where we can think about it simply: Out of these 9 remaining people, exactly 2 will be chosen (1 for secretary, 1 for treasurer). So Harry's chance of being one of these 2 chosen people is 2 out of 9, or \(\frac{2}{9}\).

Why \(\frac{2}{9}\)? Because from Harry's perspective, there are 2 "winning" spots (secretary or treasurer) out of 9 total remaining people.

4. Apply probability combination logic

Now we combine our two parts:

The probability that Harry becomes secretary or treasurer = (Probability Harry avoids presidency) × (Probability Harry gets secretary or treasurer given he avoided presidency)

In everyday terms: (Chance he's not president) × (Chance he's picked for one of the other two jobs)

Mathematically: \(\frac{9}{10} \times \frac{2}{9} = \frac{18}{90} = \frac{1}{5}\)

Notice how the 9's cancel out: \(\frac{9}{10} \times \frac{2}{9} = \frac{9 \times 2}{10 \times 9} = \frac{18}{90} = \frac{1}{5}\)

4. Final Answer

The probability that Harry will be either secretary or treasurer is \(\frac{1}{5}\).

Checking against the answer choices, this matches choice E: \(\frac{1}{5}\).

This makes intuitive sense too - Harry has a pretty decent chance (20%) of landing one of these two positions, which feels reasonable given that 2 out of 10 members will fill the roles he wants, but he first needs to avoid the presidency.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the constraint that Harry cannot hold multiple positions
Students often miss that if Harry becomes president, he's automatically eliminated from being secretary or treasurer. They might incorrectly try to calculate the probability that Harry gets any of the three positions, rather than specifically secretary OR treasurer (which requires him NOT to be president first).

2. Confusing sequential selection with simultaneous selection
The problem involves sequential selection (president chosen first from 10, then secretary from remaining 9, then treasurer from remaining 8), but students might treat this as if all three positions are filled simultaneously. This leads to incorrect probability calculations because they don't account for the changing pool of candidates.

3. Misinterpreting "either secretary or treasurer" as requiring both positions
Students might misread the question and think Harry needs to be chosen for both secretary AND treasurer roles, rather than understanding that the question asks for the probability he gets one OR the other of these two specific positions.

Errors while executing the approach

1. Calculating P(Harry avoids presidency) incorrectly
When determining that Harry has a \(\frac{9}{10}\) chance of not being president, students might incorrectly calculate this as \(\frac{1}{10}\) (confusing the probability he IS chosen as president with the probability he ISN'T chosen). This fundamental error cascades through the entire calculation.

2. Arithmetic errors in fraction multiplication and simplification
When multiplying \(\frac{9}{10} \times \frac{2}{9}\), students often make computational mistakes such as: forgetting to cancel the 9s, incorrectly multiplying numerators or denominators, or making errors when simplifying \(\frac{18}{90}\) to \(\frac{1}{5}\).

3. Incorrectly calculating Harry's chances among the remaining 9 people
Students might calculate Harry's probability as \(\frac{1}{9}\) instead of \(\frac{2}{9}\), forgetting that there are TWO favorable positions (secretary and treasurer) available among the 9 remaining candidates after the presidency is filled.

Errors while selecting the answer

1. Selecting a probability that seems "too high"
Students might second-guess themselves when they arrive at \(\frac{1}{5} = 0.2 = 20\%\), thinking this probability seems too large. They might then incorrectly choose a much smaller probability like \(\frac{1}{80}\) or \(\frac{1}{720}\), not realizing that having a 20% chance for 2 out of 10 possible positions (with constraints) is actually reasonable.