Jeevan puts the fish he took out, back into the tank. He puts some more white fish into the tank. Jeevan takes at randoun a fish from the tank. The probability that he takes a white fish is now \( \frac{1}{2} \) (b) Work out the number of white fish Jeevan put into the tank.

To solve this problem, let's define the variables and the known conditions step by step. 1. **Define Variables:** - Let \( w \) be the initial number of white fish in the tank. - Let \( r \) be the initial number of red fish in the tank. - Let \( x \) be the number of white fish Jeevan puts into the tank. 2. **After Jeevan's Actions:** - The total number of white fish in the tank becomes \( w + x \). - The total number of fish in the tank becomes \( (w + x) + r \). 3. **Probability Condition:** - The probability that Jeevan takes a white fish is given as \( \frac{1}{2} \). This can be expressed mathematically as: \[ \frac{w + x}{(w + x) + r} = \frac{1}{2} \] 4. **Cross-Multiplying:** - To eliminate the fraction, we can cross-multiply: \[ 2(w + x) = (w + x) + r \] 5. **Simplifying the Equation:** - Rearranging the equation gives: \[ 2w + 2x = w + x + r \] \[ 2w + 2x - w - x = r \] \[ w + x = r \] Now we have established a relationship between the number of white fish, the number of white fish added, and the number of red fish in the tank. 6. **Conclusion:** - The number of white fish Jeevan put into the tank is \( x = r - w \). To find the exact value of \( x \), we need additional information about either \( w \) or \( r \). If you have specific values for the initial number of white or red fish, please provide them, and we can calculate \( x \).