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AS Level Maths Probability Question Help

Can anyone help me with this 1 mark Probability Question?

Bag A contains 3 black discs and 2 white discs only. Initially Bag B is empty. Discs are removed at random from bag A, and are placed in bag B, one at a time, until all 5 discs are in bag B.
Write down the probability that the last disc that is placed in bag B is black. [1]

(the answer is 3/5 but i don't see how to get to it)
(edited 1 month ago)
Reply 1
Original post by Dhshsnsksnskxn
Can anyone help me with this 1 mark Probability Question?
Bag A contains 3 black discs and 2 white discs only. Initially Bag B is empty. Discs are removed at random from bag A, and are placed in bag B, one at a time, until all 5 discs are in bag B.
Write down the probability that the last disc that is placed in bag B is black. [1]
(the answer is 3/5 but i don't see how to get to it)

If you thought about all the possible combinations of 5 white and black discs, for any of the positions, what would be the probability of a black occurring?

Or if the discs are numbered 1..5, then theyd all be equally likely to be the last one and 3 are black so ...
(edited 1 month ago)
Original post by mqb2766
If you thought about all the possible combinations of 5 white and black discs, for any of the positions, what would be the probability of a black occurring?
Or if the discs are numbered 1..5, then theyd all be equally likely to be the last one and 3 are black so ...
if you were to pick a black disc at random then the probability would be 3/5 i get that, but for it to be specifically in the last position why doesn't it change?
Reply 3
Original post by Dhshsnsksnskxn
if you were to pick a black disc at random then the probability would be 3/5 i get that, but for it to be specifically in the last position why doesn't it change?

If you had 3 discs (maybe 2 black, say 1 and 2), then all combinations would be
1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1
If you dont know what has occurred before, then every disc is equally likely to be picked in any position. So disc 1 occurs twice in position 1, and position 2 and position 3. So disc 1 would have a probabilty of 1/3 of being picked last. Similarly for discs 2 and 3.
Original post by mqb2766
If you had 3 discs (maybe 2 black, say 1 and 2), then all combinations would be
1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1
If you dont know what has occurred before, then every disc is equally likely to be picked in any position. So disc 1 occurs twice in position 1, and position 2 and position 3. So disc 1 would have a probabilty of 1/3 of being picked last. Similarly for discs 2 and 3.

Ah i now understand what you mean, thank you

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