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.
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Let's denote the number of white fish that Jeevan originally had in the tank as \( W \), and the number of fish he took out before the new fish were added as \( R \). After putting the fish back, additional white fish \( x \) are added to the tank. Thus, the new number of white fish becomes \( W + x \) and the total number of fish in the tank also increases by \( x \), resulting in a total of \( T + x \), where \( T \) is the initial total number of fish. The probability of taking a white fish now is given by: \[ \frac{W + x}{T + x} = \frac{1}{2} \] Cross-multiplying gives: \[ 2(W + x) = T + x \] \[ 2W + 2x = T + x \] \[ 2W + x = T \] From this relationship, if we can express the initial total number of fish \( T \) in terms of \( W \) and \( x \), we can solve for \( x \). Given that \( T = W + R \) (the original number of white fish plus the number of random fish taken out), we can also equate it and solve. Assuming Jeevan started with \( W = 2 \) and \( R = 2 \), we add \( x \): Setting up the equation \( 2(2) + x = 2 + x \), we simplify to find \( x = 2 \). So, the number of white fish Jeevan put into the tank is 2!
