A bag contains X red balls , 2 white and 3 green balls. 2 balls are picked up from the bag at random, one after the othe, without replacement, the probability of being both red is 4/7. What is the value of X? एक बैग में x लाल गेंद, 2 सफेद गेंद और 3 हरी गेंद हैं। 2 गेंदों को बैग से यादृच्छिक रूप से, एक के बाद एक, प्रतिस्थापन के बिना, निकाला जाता है दोनों लाल गेंद होने की प्रायिकता 4/7 है। X का मान क्या है?

Total balls = X + 2 + 3 = X + 5. Probability of both red (without replacement) = (X/(X+5)) × ((X-1)/(X+4)) = X(X-1)/((X+5)(X+4)) = 4/7. Cross-multiplying: 7X(X-1) = 4(X+5)(X+4). This gives 7X² - 7X = 4X² + 36X + 80, so 3X² - 43X - 80 = 0. Using quadratic formula: X = (43 ± √(1849 + 960))/6 = (43 ± √2809)/6 = (43 ± 53)/6. X = 96/6 = 16 (valid) or X = -10/6 (invalid). Since 16 is not in the options, 'None of these' is correct.