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In a sports equipment allocation simulation conducted in Sydney, a coach is distributing 69 basketballs and 57 footballs to two teams using a chance experiment. The coach conducts a two-step experiment: first, a ball is randomly selected from the combined pool of 126 balls (69 basketballs and 57 footballs) and removed without replacement; second, another ball is selected from the remaining 125.
What is the exact probability that both selected balls are basketballs? Express your answer as a simplified fraction.
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Worked solution
- The total number of balls is 69 basketballs + 57 footballs = 126 balls.
- The first ball is selected at random from 126 balls. The probability it is a basketball is 69/126.
- After removing one basketball, 68 basketballs and 57 footballs remain, for a total of 125 balls.
- The probability the second ball is a basketball, given the first was a basketball, is 68/125.
- The probability of both events occurring (both basketballs) is the product of the two probabilities: (69/126) × (68/125).
- Multiply numerators: 69 × 68 = 4692.
- Multiply denominators: 126 × 125 = 15750.
- The combined fraction is 4692 / 15750.
- Simplify the fraction by finding the greatest common divisor (GCD) of 4692 and 15750. First, factor both numbers.
- GCD(4692, 15750) = 6. Divide numerator and denominator by 6: 4692 ÷ 6 = 782, 15750 ÷ 6 = 2625.
- The simplified fraction is 782 / 2625. Check that 782 and 2625 have no common factors (782 = 2×17×23, 2625 = 3×5^3×7).
- Therefore, the exact probability that both selected balls are basketballs is 782/2625.
Reference answer: 782/2625
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