Grace has 16 jellybeans in her pocket. She has 8 red ones, 4 green ones, and 4 blue ones. What is the minimum number of jellybeans she must take out of her pocket to ensure that she has one of each color? Please tell me HOW to solve this problem. Giving me the answer is somewhat useless. I know it's a PRAXIS test question and I know there are answers posted but I'm more interested in knowing what area of probability I have to bone up on. Thank you one and all.
2006-11-16 11:29:02 · 9 answers · asked by jbowers9 1 in Education & Reference ➔ Trivia
This is really not a probability problem but a logic problem.
You ask what the minimum number of beans she must take
out of her pocket, to ensure that she has one of each color.
Sometimes it is best to turn the question around and work backwards.
Suppose I told you, "You can choose any marble from my pocket you like, you choose the color. What is the Most
marbles you can pull out and Not have one of each color?"
Well, first you would pull out 8 red ones, then you could take 4
green ones, All that are left are blue ones, so when you pull out the 13th marble, you have to have one of each color.
When you come up with an answer, all you have to do is find
one case which refutes your answer to tell you it is wrong.
Say I told you that the minimum was 10 marbles. You could refute that by showing me one case where you took out 10 marbles and STILL did not have one of each color.
Use the case where you took out 8 red and 2 green and still
had no blue.
But you can find no case where you can take out 13 marbles
and not have one of each.
Hope this helps you. Remember to reason the problem out before plugging in numbers.
2006-11-16 12:14:42 · answer #1 · answered by True Blue 6 · 3⤊ 0⤋
Well, I assume that she can't see what she's selecting... so the answer's going to be 13: naturally, she's likely to need less, but the worst case scenario is that she draws out 8 red ones, and then either 4 green or 4 blue ones for 12... the next one will be the third and final color. Drawing 13 is the only way to ensure that she'll have at least one of all three colors.
2006-11-16 11:38:50 · answer #2 · answered by retfordt 2 · 3⤊ 0⤋
Green Jelly Beans
2016-12-18 17:51:54 · answer #3 · answered by Anonymous · 0⤊ 0⤋
She is unlucky and the first eight she picks are red.
So she carries on and she picks another four and guess what!
She picked the green ones!
So, she picks another one and...YES its a blue one!
At last! She has one of each color.
So 8+4+1=13.
Unlucky number.
Really unlucky Grace with teeth full of cavities!
2006-11-20 02:06:45 · answer #4 · answered by I Drum 3 · 1⤊ 0⤋
it is a ratio problem. the highest possibilty is that she draws 8 red, 4 green and 1 blue. that is 13, the high. the lowest is choosing 1 red, 1 green, and 1 blue. that is three, so the ratio is
low to high
(3 : 13)
2006-11-16 11:42:19 · answer #5 · answered by conƒused-little-man 2 · 2⤊ 1⤋
Yeah, it's thirteen. She could take eight in a row and get all red, and then four more and get all blue.
2006-11-16 12:34:35 · answer #6 · answered by Anonymous · 0⤊ 1⤋
The minimum or least things she could take out is three since there are three colors you take out three things to know how to get the three colors.
2006-11-16 12:26:14 · answer #7 · answered by . 2 · 0⤊ 3⤋
The answer is three. If she has three COLORS of beans and she has to take out one of EACH COLOR she needs to take out 3. They didn't ask probability.
2006-11-16 11:34:26 · answer #8 · answered by Anonymous · 0⤊ 3⤋
dude, you can't even spell practice right!!!
2006-11-16 12:13:12 · answer #9 · answered by yuuki chan 3 · 0⤊ 5⤋