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.
(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!
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.
To solve this problem, we need to use the concept of conditional probability. Let's break down the problem into parts:
(a) Probability of Andy drawing a white disc on his second turn:
To determine the probability, we need to consider the outcomes of the previous turns. Andy already drew a white disc on his first turn. Now, there are 4 white and 5 black discs left in the bag. Thus, the probability of Andy drawing a white disc on his second turn is 4/9.
(b) Probability of Andy failing to draw a white disc on his second turn but Beth drawing a black disc on her second turn:
Again, we need to consider the outcomes of the previous turns. Andy drew a white disc on his first turn, so there are 4 white and 5 black discs left. If Andy fails to draw a white disc, it means he draws a black disc. Now, there are 4 white and 4 black discs remaining. The probability of Beth drawing a black disc on her second turn, given that Andy drew a black disc on his second turn, is 4/8 or 1/2.
(c) Probability that after three draws, Andy has two discs of the same color:
To determine this probability, we need to consider the different possibilities for the outcomes of the three draws. There are two ways for Andy to end up with two discs of the same color: either he draws two white discs or two black discs.
(i) Probability of Andy drawing two white discs:
The probability of Andy drawing a white disc on his first turn is 5/10. After that, there are 4 white and 5 black discs left. The probability of drawing a second white disc is 4/9. So, the probability of drawing two white discs is (5/10) * (4/9) = 20/90 = 2/9.
(ii) Probability of Andy drawing two black discs:
The probability of Andy drawing a black disc on his first turn is 5/10. After that, there are 5 white and 4 black discs left. The probability of drawing a second black disc is 4/9. So, the probability of drawing two black discs is (5/10) * (4/9) = 20/90 = 2/9.
Since these are the only two possibilities, we can add the probabilities together: 2/9 + 2/9 = 4/9.
(b) Beth's chance of drawing a black disc before Andy's second turn:
To calculate this probability, again we consider the different possibilities:
(i) Beth draws a black disc on her first turn, and Andy draws a white disc on his first turn:
The probability of Beth drawing a black disc on her first turn is 5/10. After that, there are 5 white and 4 black discs left in the bag. The probability of Andy drawing a white disc on his first turn is 5/9. So, the probability is (5/10) * (5/9) = 25/90 = 5/18.
(ii) Beth draws a white disc on her first turn, and Andy draws a black disc on his first turn:
The probability of Beth drawing a white disc on her first turn is 5/10. After that, there are 4 white and 5 black discs left in the bag. The probability of Andy drawing a black disc on his first turn is 5/9. So, the probability is (5/10) * (5/9) = 25/90 = 5/18.
Since these are the only two possibilities, we can add the probabilities together: 5/18 + 5/18 = 10/18 = 5/9.
Therefore, Beth's chance of drawing a black disc before Andy's second turn is 5/9.