Probability

knight247 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?
1/4
1/2
1/2
5/8
2/3
3/4

Don't have an OA. Detailed explanations would be appreciated. Thanks

YBBB; probability that 4th ball is black = 3/5 (as we have already picked 2 black balls, so the remaining are 3 balls)
YYBB; probability that 4th ball is black = 4/5 (as we have already picked 1 black ball, so the remaining are 4 balls)
YYYB; probability that 4th ball is black = 5/5 (as we have not picked any black ball, so there are 5 balls)
BBBB; probability that 4th ball is black = 2/5 (as we have already picked 3 black balls, so the remaining are 2 balls)

Therefore required probability = [(3C1 * 5C2)/8C3] * 3/5 + [(3C2 * 5C1)/8C3] * 2/5 + [3C3/8C3] * 1 + (5C3/8C3) * 2/5 = (30/56) * (3/5) + (15/56) * (4/5) + (1/56) + (10/56) * (2/5) = (18 + 12 + 1 + 4)/56 = 35/56 = [spoiler]5/8[/spoiler]

shankar.ashwin wrote:Also to add,
P(B) on any given pick = P(B) on the first pick.
Hence 5/8

Exactly. Nice work, shankar.ashwin!

This reminds me of when I was young, and my friends and I would sometimes "draw straws" to randomly select one person for a task (usually a less than ideal task like getting wood for the fire, or something stupid 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