Solve the problem using three (numerical) techniques: 1. Using a for loop with an if() clause and accumulator (Hint: The strategy is to sample 3 fishes without replacement from sample space, sample space <- c(rep("Y",3),rep("B",7)) n <- 1000000 and count how many fishes were yellow. If only one is yellow, then this is a successful case. In this case, add 1 to the accumulator. Repeat the procedure n times. The frequentist result will be accumulator/n.
2. Using a for loop with an if else() function and accumulator.
3. Vectorial approach using sapply (no loops, no if() or ifelse(), no accumulator).