A box contains 5 red balls and 3 green balls. Two balls are drawn at random without replacement. What is the probability that both balls are red?
Show Hint
- $ \frac{1}{7} $
- $ \frac{5}{28} $
- $ \frac{5}{14} $
- $ \frac{10}{28} $
The Correct Option is C
Solution and Explanation
Since the drawing is without replacement, combinations are used to calculate probabilities.
Total number of ways to choose 2 balls from 8: \[ \binom{8}{2} = \frac{8 \cdot 7}{2} = 28 \]
Number of favorable outcomes (choosing 2 red balls from 5): \[ \binom{5}{2} = \frac{5 \cdot 4}{2} = 10 \]
Probability that both balls are red: \[ \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{10}{28} = \frac{5}{14} \]
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Based upon the results of regular medical check-ups in a hospital, it was found that out of 1000 people, 700 were very healthy, 200 maintained average health and 100 had a poor health record.
Let \( A_1 \): People with good health,
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