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Can someone please solve this ASAP!!! (1 Viewer)
- Thread starter Jayc2112
- Start date
(i) P(3 red) = 25/75 * 24/74 * 23/73
Clearly, the ticket that wins the first draw has a 1/3 chance of being red. When that ticket is removed there are 24 red tickets, 25 blue and 25 green, making 74 tickets in total. Therefore the next ticket drawn has a 24/74 chance of being red. After the second ticket is drawn there are 23 red tickets, 25 blue and 25 green for a total of 73, so the probability of another red ticket is 23/73.
(ii) P(1 red, 1 green, 1 blue) = 1 * 50/74 * 25/73
The colour of the first ticket drawn does not matter. After that ticket is drawn, there are 24 of that colour and 25 each of the other colours, making 74 in total. The next ticket drawn can be either of those other colours (as long as it wasn't the same colour as the first), so the probability is 50/74. Finally, after the second ticket is drawn, there are 73 remaining tickets, 24 of two colours and 25 of another. We want the final colour, the one that hasn't had a ticket drawn yet, so the probability is 25/73.
Clearly, the ticket that wins the first draw has a 1/3 chance of being red. When that ticket is removed there are 24 red tickets, 25 blue and 25 green, making 74 tickets in total. Therefore the next ticket drawn has a 24/74 chance of being red. After the second ticket is drawn there are 23 red tickets, 25 blue and 25 green for a total of 73, so the probability of another red ticket is 23/73.
(ii) P(1 red, 1 green, 1 blue) = 1 * 50/74 * 25/73
The colour of the first ticket drawn does not matter. After that ticket is drawn, there are 24 of that colour and 25 each of the other colours, making 74 in total. The next ticket drawn can be either of those other colours (as long as it wasn't the same colour as the first), so the probability is 50/74. Finally, after the second ticket is drawn, there are 73 remaining tickets, 24 of two colours and 25 of another. We want the final colour, the one that hasn't had a ticket drawn yet, so the probability is 25/73.