Question

a bag contain 24 balls out of which x are white .if one ball drawn at random ,the probability of drawing white ball is "y".12 more white balls are added to the bag .now if a ball is drawn from the probability of drawing white ball is "5/3y" .find the value of x?

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Probability}\mathrm{of}\mathrm{drawing}\mathrm{a}\mathrm{white}\mathrm{ball}=\frac{\mathrm{number}\mathrm{of}\mathrm{white}\mathrm{ball}}{\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{balls}} \\ \mathbf{Given}\mathbf{:}\mathrm{y}=\frac{\mathrm{x}}{24}....\left(1\right) \\ \mathrm{According}\mathrm{to}\mathrm{question},\mathrm{when}12\mathrm{more}\mathrm{white}\mathrm{balls}\mathrm{are}\mathrm{added}, \\ \mathrm{probability}\mathrm{of}\mathrm{drawing}\mathrm{a}\mathrm{white}\mathrm{ball}=\frac{5\mathrm{y}}{3} \\ \mathrm{So},\frac{5\mathrm{y}}{3}=\frac{\mathrm{x}+12}{24+12} \\ \frac{5\mathrm{y}}{3}=\frac{\mathrm{x}+12}{36} \\ 5\mathrm{y}=\frac{\mathrm{x}+12}{12} \\ \mathrm{substituting}\mathrm{value}\mathrm{of}\mathrm{y}\mathrm{from}\mathrm{equ}.\left(1\right) \\ 5\times \frac{\mathrm{x}}{24}=\frac{\mathrm{x}+12}{12} \\ \frac{5\mathrm{x}}{2}=\mathrm{x}+12 \\ 5\mathrm{x}=2\mathrm{x}+24 \\ 3\mathrm{x}=24 \\ \mathrm{x}=\frac{24}{3}=8 \end{gathered}$$

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