The probability that event M will not occur is 0.8 and the probability that event R will not occur is...

1. Translate the problem requirements: Convert "will not occur" probabilities to "will occur" probabilities, and clarify that "cannot both occur" means the events are mutually exclusive
2. Calculate individual event probabilities: Find \(\mathrm{P(M)}\) and \(\mathrm{P(R)}\) from the given complement probabilities
3. Apply mutually exclusive events rule: Since events cannot both occur, use the simplified addition rule \(\mathrm{P(M \text{ or } R)} = \mathrm{P(M)} + \mathrm{P(R)}\)
4. Verify against answer choices: Confirm the calculated probability matches one of the given options

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what the problem is telling us in everyday language:

• We're told "the probability that event M will not occur is 0.8" - this means event M has an 80% chance of NOT happening
• We're told "the probability that event R will not occur is 0.6" - this means event R has a 60% chance of NOT happening
• The phrase "events M and R cannot both occur" means these events are mutually exclusive - like flipping a coin and getting both heads and tails at the same time, which is impossible

Now we need to convert these "will not occur" probabilities into "will occur" probabilities. Think of it this way: if there's an 80% chance something won't happen, then there's a 20% chance it will happen.

Process Skill: TRANSLATE - Converting problem language about complements and mutual exclusivity into mathematical understanding

2. Calculate individual event probabilities

Since probabilities of an event occurring and not occurring must add up to 1 (or 100%), we can find:

• If \(\mathrm{P(M \text{ will not occur})} = 0.8\), then \(\mathrm{P(M \text{ will occur})} = 1 - 0.8 = 0.2\)
• If \(\mathrm{P(R \text{ will not occur})} = 0.6\), then \(\mathrm{P(R \text{ will occur})} = 1 - 0.6 = 0.4\)

In mathematical notation: \(\mathrm{P(M)} = 0.2\) and \(\mathrm{P(R)} = 0.4\)

Let's convert these to fractions to match our answer choices:

• \(\mathrm{P(M)} = 0.2 = \frac{2}{10} = \frac{1}{5}\)
• \(\mathrm{P(R)} = 0.4 = \frac{4}{10} = \frac{2}{5}\)

3. Apply mutually exclusive events rule

Since events M and R cannot both occur (they're mutually exclusive), finding the probability that "either event M or event R will occur" becomes much simpler.

Think of it like this: imagine you have a bag with 10 marbles. Event M happens with 2 of those marbles, and event R happens with 4 of those marbles. Since M and R can't both happen, these are completely different marbles. So "either M or R" happens with 2 + 4 = 6 marbles out of 10.

For mutually exclusive events, we simply add the individual probabilities:
\(\mathrm{P(M \text{ or } R)} = \mathrm{P(M)} + \mathrm{P(R)} = \frac{1}{5} + \frac{2}{5} = \frac{3}{5}\)

Process Skill: APPLY CONSTRAINTS - Using the mutual exclusivity condition to simplify the calculation

4. Verify against answer choices

Our calculated probability is \(\frac{3}{5}\).

Looking at the answer choices:

1. \(\frac{1}{5}\)
2. \(\frac{2}{5}\)
3. \(\frac{3}{5}\) ← This matches our answer
4. \(\frac{4}{5}\)
5. \(\frac{12}{25}\)

Let's double-check: \(\frac{3}{5} = 0.6\), which means there's a 60% chance that either event M or event R will occur. This makes sense given our individual probabilities of 20% and 40%.

Final Answer

The probability that either event M or event R will occur is \(\frac{3}{5}\).

The correct answer is C) \(\frac{3}{5}\).

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "cannot both occur" constraint

Students often overlook or misunderstand the phrase "events M and R cannot both occur." They may proceed to use the general formula \(\mathrm{P(M \text{ or } R)} = \mathrm{P(M)} + \mathrm{P(R)} - \mathrm{P(M \text{ and } R)}\) without recognizing that the mutual exclusivity means \(\mathrm{P(M \text{ and } R)} = 0\). This leads them to think they need additional information to solve the problem.

2. Working directly with complement probabilities

Some students attempt to work directly with the given "will not occur" probabilities (0.8 and 0.6) without first converting them to "will occur" probabilities. They might try to subtract these from 1 or combine them in incorrect ways, missing the fundamental step of finding \(\mathrm{P(M)} = 0.2\) and \(\mathrm{P(R)} = 0.4\).

3. Confusing "either...or" with "exactly one"

Students may misinterpret "either event M or event R will occur" as meaning "exactly one of M or R occurs" rather than "at least one of M or R occurs." While this doesn't affect the final answer in this particular problem due to mutual exclusivity, the conceptual misunderstanding can lead to confusion in the approach.

Errors while executing the approach

1. Arithmetic errors in fraction addition

When adding \(\mathrm{P(M)} + \mathrm{P(R)} = \frac{1}{5} + \frac{2}{5}\), students may make basic arithmetic mistakes, such as adding numerators and denominators separately (getting \(\frac{3}{10}\) instead of \(\frac{3}{5}\)) or incorrectly converting decimals to fractions.

2. Incorrect complement calculation

Students might incorrectly calculate the individual probabilities by using \(\mathrm{P(M)} = 0.8\) instead of \(\mathrm{P(M)} = 1 - 0.8 = 0.2\), essentially using the given "will not occur" probabilities as "will occur" probabilities.

Errors while selecting the answer

1. Selecting decimal equivalent without checking fraction form

Students who correctly calculate 0.6 as their final answer might look for this decimal form among the choices rather than converting to the fraction \(\frac{3}{5}\). This could lead them to incorrectly select choice E) \(\frac{12}{25}\) (which equals 0.48) if they misread or miscalculate.

2. Choosing individual probability instead of combined probability

After correctly calculating \(\mathrm{P(M)} = \frac{1}{5}\) and \(\mathrm{P(R)} = \frac{2}{5}\), some students might accidentally select one of these individual probabilities (choice A or B) instead of their sum, especially if they lose track of what the question is actually asking for.