If a rubber ball consistently bounces back $$\frac{{2}}{{3}}$$ of the height from which it is dropped, what fraction of its original height will the ball bounce after being dropped and bounced four times without being stopped?

Solution(By Examveda Team)

Each time the ball is dropped and it bounces back, it reaches $$\frac{{2}}{{3}}$$ of the height it was dropped from.
After the first bounce, the ball will reach $$\frac{{2}}{{3}}$$ of the height from which it was dropped - let us call it the original height. After the second bounce, the ball will reach $$\frac{{2}}{{3}}$$ of the height it would have reached after the first bounce.
So, at the end of the second bounce, the ball would have reached $$\frac{{2}}{{3}}$$ × $$\frac{{2}}{{3}}$$ of the original height = $$\frac{{4}}{{9}}$$ th of the original height.
After the third bounce, the ball will reach $$\frac{{2}}{{3}}$$ of the height it would have reached after the second bounce.
So, at the end of the third bounce, the ball would have reached $$\frac{{2}}{{3}}$$ × $$\frac{{2}}{{3}}$$ × $$\frac{{2}}{{3}}$$ = $$\frac{{8}}{{27}}$$ th of the original height.
After the fourth and last bounce, the ball will reach $$\frac{{2}}{{3}}$$ of the height it would have reached after the third bounce.
So, at the end of the last bounce, the ball would have reached $$\frac{{2}}{{3}}$$ × $$\frac{{2}}{{3}}$$ × $$\frac{{2}}{{3}}$$ × $$\frac{{2}}{{3}}$$ of the original height = $$\frac{{16}}{{81}}$$ of the original height.