A bag contains 5 white discs and 5 black discs. Two players draw discs from the bag one at a time in turns and do not replace them. Andy goes first and draws a white disc. Beth follows and draws a white disc.

Question

(a)What is the probability:
(i)Andy draws a white disc on his second turn
(ii)Andy fails to draw a white disc on his second turn but Beth draws a black disc on her second turn
(iii)after three dips into the bag Andy has two discs of the same colour?
(b)What is Beth's chance of drawing a black disc if she has her second go before Andy has his second go?

Really confused so any help, thanks in advance!

Answer 1

There are only 10 disks. Once the two white disks are drawn, 3 white and 5 black remain.

i. What is the proportion of white disks to the remaining total for Andy?

ii. This means that both draw black disks. For Andy, it is the proportion of black disks to the remaining total. For Beth, there are 4 black disks remaining out of the remaining 7. The probability of both/all events is found by multiplying the probability of the individual events.

iii. Andy already has a white disk, so you are seeking the probability of getting another white disk on either the second or third pick or two black disks in a row. However, you need to remember that Beth has picks in between. Either-or probabilities are obtained by adding the individual probabilities. This gets complex.

b. It would be the same as Andy drawing a black disk on his second turn. (iia)

I hope this helps. Thanks for asking.