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Three seeds each germinate with probability $\tfrac13$. The probability that at least two germinate is:

A$\tfrac{1}{27}$
B$\tfrac{7}{27}$
C$\tfrac{6}{27}$
D$\tfrac{8}{27}$
Answer & Solution
Correct answer: B. $\tfrac{7}{27}$
1. 'At least two' covers exactly two germinating and all three germinating. 2. Exactly three: $\left(\tfrac13\right)^3=\tfrac1{27}$. 3. Exactly two: choose which two in 3 ways, giving $3\times\left(\tfrac13\right)^2\times\tfrac23=\tfrac6{27}$. 4. The two cases are mutually exclusive, so add them. 5. Total $=\tfrac6{27}+\tfrac1{27}=\tfrac7{27}$. Stopping at $\tfrac6{27}$ forgets the all-three branch of the tree diagram. _Source: NECTA CSEE 2022 Basic Mathematics 041, Question 3: Sets and Probability_
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