Roger.Robert
New member
- Joined
- Jan 27, 2018
- Messages
- 30
In a pencil box there are 8 crayons:3 red,2 blue and 3 black.If we take 5 crayons from the pencil box, how many possibilities are there so that at least two crayons are red?
You should know by now that we want to see what work you have done or at least hear what ideas you have had.In a pencil box there are 8 crayons:3 red,2 blue and 3 black.If we take 5 crayons from the pencil box, how many possibilities are there so that at least two crayons are red?
If you took at least two red crayons, then how many red crayons might you have taken? Show us your work!In a pencil box there are 8 crayons:3 red,2 blue and 3 black.If we take 5 crayons from the pencil box, how many possibilities are there so that at least two crayons are red?
That's badly worded. Should be:In a pencil box there are 8 crayons:3 red,2 blue and 3 black.If we take 5 crayons from the pencil box, how many possibilities are there so that at least two crayons are red?
That's badly worded. Should be:
if 5 crayons are taken AT RANDOM,
what is the PROBABILITY that at least 2 crayons are red?
Are you a student attending math classes?
I think the final answer is 6.If I choose 2 red pencils I can make 3 combinations with the rest of the pencils.If I choose 3 red pencils I also can choose 3 combinations.So in total 6.
But I was wondering what these types of problems are called(where there are a limited number of identical objects,and count the number of combinations after picking n number of objects), so I can find the general methods to solve.
Here is a completely different way to approach this problem.I think the final answer is 6.If I choose 2 red pencils I can make 3 combinations with the rest of the pencils.If I choose 3 red pencils I also can choose 3 combinations.So in total 6.
But I was wondering what these types of problems are called(where there are a limited number of identical objects,and count the number of combinations after picking n number of objects), so I can find the general methods to solve.