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Probability of winning prizes in boxes?
A TV has a game show in which there are nine boxes, some of which contain prizes. Three of the boxes are yellow, three are red and three are blue. For each colour the three boxes are numbered from 1 to 3.
In the game the contestant is asked to remove all the boxes with two of the three numbers and one of the three colours. For example, remove all of the boxes labelled 1 and 2 and then remove the remaining box labelled blue. The contestant then wins any prizes found in the boxes not removed.
Q1. The game is expanded by adding a fourth colour, green, and having four boxes of each colour, each labelled 1 to 4. The rules now are that the contestant nominates two colours and two numbers to be totally eliminated. What is the minimum number of boxes into which the TV station must put prizes to ensure that there is at least one remaining box containing a prize? Explain your answer.
3 Answers
- 1 decade agoFavorite Answer
Therefore, 12 boxes will have to be selected and removed leaving 4 boxes remaining. When the TV station must ensure that at least one remaining prize is available, they must add 1 to half of 16 so that anwer is nine
Think of it as if you removed green and blue
there goes numbers 1 2 3 and 4 for both colours. That's eight gone. Then think of two numbers to be removed, say 1 and 2. Since the blue and green are already removed, you only have yellow and red to remove. That makes four boxes gone from red and yellow. Add those up, 8 + 4 =12
There are 16 boxes so the difference is four remaining. Since to be sure that there needs to be a prize in one of the boxes, you must add one prize greater than half of 16, which is 9.
- 5 years ago
Probability of getting the right box = 1/3 Probability of getting the prize given the right box = 1/2 Probability of getting the prize = (1/3)(1/2) = 1/6
- Anonymous1 decade ago
HAHAHAH
This was one of the questions out of the Australian Mathematics Competition for High Schools?
But it was due in about 3 weeks ago...