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Question:
In a box there are (x + 5) red balls, ‘x’ blue balls and (x + 1) black balls. If the probability of picking two blue balls at random is 1/35, then find the value of ‘x’.
In a box there are (x + 5) red balls, ‘x’ blue balls and (x + 1) black balls. If the probability of picking two blue balls at random is 1/35, then find the value of ‘x’.
Updated On: Sep 13, 2024
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The Correct Option is D
Solution and Explanation
According to the question,
Total number of blue balls in the bag = (x + 5 + x + x + 1) = (3x + 6)
According to the question,
\(\left\{{^xC_2}/{^{(3x+6)}C_2}\right\}=\frac{1}{35}\)
\(Or, \frac{x(x – 1)}{(3x + 6)(3x + 5)} = \frac{1}{35}\)
Or, 35x² – 35x = 9x² + 33x + 30
Or, 26x² – 68x – 30 = 0
Or, 26x² – 78x + 10x – 30 = 0
Or, 26x(x – 3) + 10(x – 3) = 0
Or, (26x + 10)(x – 3) = 0
Since, number of balls cannot be negative
Therefore, x = 3
So, the correct option is (D) : 3.
Total number of blue balls in the bag = (x + 5 + x + x + 1) = (3x + 6)
According to the question,
\(\left\{{^xC_2}/{^{(3x+6)}C_2}\right\}=\frac{1}{35}\)
\(Or, \frac{x(x – 1)}{(3x + 6)(3x + 5)} = \frac{1}{35}\)
Or, 35x² – 35x = 9x² + 33x + 30
Or, 26x² – 68x – 30 = 0
Or, 26x² – 78x + 10x – 30 = 0
Or, 26x(x – 3) + 10(x – 3) = 0
Or, (26x + 10)(x – 3) = 0
Since, number of balls cannot be negative
Therefore, x = 3
So, the correct option is (D) : 3.
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