A week ago, I helped Brian get his Christmas tree out and decorate it. He's very picky, so only wants exactly 2 baubles of each colour on the tree. He keeps all of his baubles in a box, which contains 10 reds, 8 blues, 6 yellows, 4 greens and 2 purples. It's dark, so I can't see the colour of the baubles. How many baubles do I have to pick out of the box in order to definitely meet Brian's requirements?
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This type of problem requires us to think of the worst case scenario. The worst case scenario is necessary because any other scenario may not always meet Brain's requirements. The worst case scenario is that we pick out all the reds, then all the blues, all the yellows, etc. This is the worst case scenario because it requires us to pick up the most baubles - 3 0 . Any other scenario would not ensure that we meet Brain's requirements. For example, if we picked 20 baubles, we could get 10 reds, 8 blues and 2 purples, which does not meet the requirements.
Just think about doing this in real life. If you only pick 29 of the baubles, you could get 10 reds, 8 blues, 6 yellows, 4 greens, and 1 purple, which wouldn't satisfy the requirements.
You have to pick all 30 baubles.
Exactly As I Supposed The Case!
actually its an easy ,but bogus sum.....
What just happened, this was so easy I thought it couldn't be the answer
If you were to unfortunately get all 10 red baubles first, then all 8 blue ones, then all 6 yellows, then all 4 greens, and then finally both purples, you would only then meet the requirements of having 2 of each once you finally pick the last bauble. Therefore you must pick out 1 0 + 8 + 6 + 4 + 2 = 3 0 baubles in order to definitely meet Brian's requirements.
using pigeonhole principle, we find out that to pick out 2 balls of any colour we have to pick out six balls considering the worst case scenario in which first 5 is of 5 different colors the 6th one will definitely be of one of the 5 colors ensuring that 2 balls of one colour is drawn, now we need 2 balls of each color so we continue the same process 5 times which is 5*6=30 balls....please correct me if my method is wrong...
To be sure of selecting two purple baubles, you must take all of the baubles out of the box, for a total of 1 0 + 8 + 6 + 4 + 2 = 3 0 baubles.
use worst case scenario tactic.
Because there are only 2 purples, you have to pick out all of them to be sure that you get those two. Therefore it is 10 plus 8 plus 6 plus 4 plus 2, 30.
If you bring out a number of baubles less than 30 (which is everything), there's always a chance that you didn't bring out enough purples, because of their quantity of only 2.
For example, if you only brought out 29 baubles, there's still a chance that you got every single color, except one purple bauble.
Hence the answer's 3 0 .
There are only 2 purple baubles. In the worst case scenario, those 2 would be the last 2 baubles you choose, and they're the only 2 purple baubles. So, the minimum number of baubles you'd have to pick would be all of them.
To sum it all up, 1 0 + 8 + 6 + 4 + 2 = 3 0 .
If I were you, I'd get some lights on. It really isn't worth the trouble.
The key word here is "definitely." To definitely meet Brian's requirements, all baubles must be removed from the box. The sum of all the colored baubles is 30.
using pigeonhole principle, we find out that to pick out 2 balls of any colour we have to pick out six balls considering the worst case scenario in which first 5 is of 5 different colors the 6th one will definitely be of one of the 5 colors ensuring that 2 balls of one colour is drawn, now we need 2 balls of each color so we continue the same process 5 times which is 5*6=30 balls....please correct me if my method is wrong...
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Since there are only 2 purples, she has to pick out every single bauble as she could pick a purple last (then she would have 1 purple until the last bauble is selected). Therefore, the answer is 1 0 + 8 + 6 + 4 + 2 = 3 0