Question

Let 𝐸₁, 𝐸₂ and 𝐸₃ be three events such that 𝑃(𝐸₁∩𝐸₂ )=\(\frac{1}{4}\) , 𝑃(𝐸₁ ∩ 𝐸₃ )=𝑃(𝐸₂ ∩ 𝐸₃ )=\(\frac{1}{5}\) and 𝑃(𝐸₁ ∩ 𝐸₂ ∩ 𝐸₃ ) =\(\frac{1}{6}\) . Then, among the events 𝐸₁ 𝐸₂ and 𝐸₃, the probability that at least two events occur, equals

• A. \(\frac{17}{60}\)
• B. \(\frac{23}{60}\)
• C. \(\frac{19}{60}\)
• D. \(\frac{29}{60}\)

Answer 1

The Correct Option is C

The problem requires calculating the probability that at least two of the events \( E_1 \), \( E_2 \), and \( E_3 \) occur. We have the following given probabilities:

• \( P(E_1 \cap E_2) = \frac{1}{4} \)
• \( P(E_1 \cap E_3) = \frac{1}{5} \)
• \( P(E_2 \cap E_3) = \frac{1}{5} \)
• \( P(E_1 \cap E_2 \cap E_3) = \frac{1}{6} \)

To find the probability that at least two events occur, we calculate:

\(P(\text{at least two events occur}) = P(E_1 \cap E_2) + P(E_1 \cap E_3) + P(E_2 \cap E_3) - 2P(E_1 \cap E_2 \cap E_3)\) 

Substituting the given probabilities:

\(P(\text{at least two events occur}) = \frac{1}{4} + \frac{1}{5} + \frac{1}{5} - 2 \times \frac{1}{6}\)

First, we calculate each step separately:

• \(\frac{1}{4} = \frac{15}{60}\)
• \(\frac{1}{5} = \frac{12}{60}\) (twice, for \( E_1 \cap E_3 \) and \( E_2 \cap E_3 \))
• \(2 \times \frac{1}{6} = \frac{2}{6} = \frac{20}{60}\)

Now substituting these into the equation:

\(P(\text{at least two events occur}) = \frac{15}{60} + \frac{12}{60} + \frac{12}{60} - \frac{20}{60}\)

Performing the arithmetic:

\(\frac{15 + 12 + 12 - 20}{60} = \frac{19}{60}\)

Therefore, the probability that at least two events occur is \(\frac{19}{60}\).

Thus, the correct answer is \(\frac{19}{60}\).