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दो गेंद एक बॉक्स से बिना प्रतिस्थापित किए निकाली जाती है। बॉक्स में $10$ काली और $8$ लाल गेदें हैं तो प्रायिकता ज्ञात कीजिए एक काली तथा दूसरी लाल हो।
$\frac{40}{81}$
$\frac{40}{81}$
$\frac{40}{81}$
$\frac{40}{81}$
Solution
Total number of balls $=18$
Number of red balls $=8$
Number of black balls $=10$
Probability of getting first ball as red $=\frac{8}{18}=\frac{4}{9}$
The ball is replaced after the first draw.
Probability of getting second ball as black $=\frac{10}{18}=\frac{5}{9}$
Therefore, probability of getting first ball as black and second ball as red $=\frac{4}{9} \times \frac{5}{9}=\frac{20}{81}$
Therefore, probability that one of them is black and other is red
$=$ Probability of getting first ball black and second as red $+$ Probability of getting first ball red and second ball black
$=\frac{20}{81}+\frac{20}{81}$
$=\frac{40}{81}$