Question

The probability that it will rain tomorrow in cities A, B and C is \(60\%\), \(70\%\) and \(80\%\) respectively. Assuming these events are independent, the probability that it will rain tomorrow in at least one of the cities is:

• A. \(\frac{3}{250}\)
• B. \(\frac{244}{250}\)
• C. \(1\)
• D. \(\frac{9}{10}\)

Hint:

Whenever a probability question asks for “at least one”, it is usually much faster to compute \(1 - P(\text{none})\) than to add up all the individual combinations where it rains in exactly one, exactly two, or all three cities.

Answer 1

The Correct Option is B

Concept: Let \(A\), \(B\), and \(C\) represent the independent events that it rains in cities A, B, and C respectively.
• The phrase “at least one” is best calculated by looking at the complement event: the probability that it does not rain in any of the three cities tomorrow.
• The probability formula using complements is: \[ P(\text{At least one}) = 1 - P(\text{None}) = 1 - P(A' \cap B' \cap C') \]
• Since the events are independent, their complement events are also independent, allowing us to multiply their probabilities directly: \[ P(A' \cap B' \cap C') = P(A') \cdot P(B') \cdot P(C') \]

Step 1: Convert given percentage values to fraction probabilities
Let us write out the given rain probabilities as simplified fractions: \[ P(A) = 60\% = \frac{60}{100} = \frac{3}{5} \] \[ P(B) = 70\% = \frac{70}{100} = \frac{7}{10} \] \[ P(C) = 80\% = \frac{80}{100} = \frac{4}{5} \]

Step 2: Calculate the complement probabilities (no rain)
Using \(P(E') = 1 - P(E)\), find the probability of no rain for each individual city: \[ P(A') = 1 - \frac{3}{5} = \frac{2}{5} \] \[ P(B') = 1 - \frac{7}{10} = \frac{3}{10} \] \[ P(C') = 1 - \frac{4}{5} = \frac{1}{5} \]

Step 3: Compute the combined probability of no rain in any city
Multiply these independent complement probabilities together: \[ P(\text{None}) = P(A') \cdot P(B') \cdot P(C') = \frac{2}{5} \times \frac{3}{10} \times \frac{1}{5} \] Multiplying the numerators and denominators: \[ P(\text{None}) = \frac{2 \times 3 \times 1}{5 \times 10 \times 5} = \frac{6}{250} \]

Step 4: Compute the final probability of rain in at least one city
Subtract this complement result from $1$: \[ P(\text{At least one}) = 1 - \frac{6}{250} \] Finding a common denominator to subtract the values: \[ P(\text{At least one}) = \frac{250 - 6}{250} = \frac{244}{250} \] This matches option (B).