hey_thr67 wrote:A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?

As we are selecting the balls without replacement, the probability that the fourth ball selected is black is same as the probability that the first ball selected is black.

This is because, before selecting the fourth ball we have 3 yellow balls and 5 black balls in the box.

Hence, required probability = (number of black balls)/(number of total balls) = 5/8

The correct answer is C.

This question reminds me of when I was young, and my friends and I would sometimes "draw straws" to randomly select one person for some unpleasant task (like getting wood for the fire, or something dangerous, like jumping off a roof).

So, someone would hold up n pieces of grass (for n guys), and one of those pieces of grass was very short. The person who selected the shortest piece was the one to do the task.

There was always one guy who wanted to choose his piece last. His reasoning was that his chances of drawing the shortest piece was minimized since every person before him had a chance of drawing the short piece before it got to his turn.

The truth of the matter is that all n guys have a 1/n chance of selecting the shortest piece, regardless of when they select.

The same applies to the original question here.

Cheers,
Brent