Grace has 16 jellybeans in her pocket. She has 8 red ones, 4

Question

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?

A. 4
B. 8
C. 12
D. 13
E. 16

A. 4 
B. 8 
C. 12 
D. 13 
E. 16

Bunuel · Accepted Answer

The worst scenario is when she takes 8 red and 4 green, total of 12 jellybeans, and still doesn't have 3 distinct colors. But the next draw (13th) will surely be the third color jellybean (blue) as there is no other color jellybeans left in pocket.

Answer: D.

actleader · Answer

Maria,
I doubt if it is al-gebra.
It is a probability problem.

Grace, in the worst case may take out 8 red jellyes ► 4 green ► and the next will be the blue one.
So 13 - is the minimum number of jellybeans.

rhine29388 · Answer

The worst case arises when she takes 8 red and 4 green, and still doesn't have 3 distinct colors. So she draws (13th) next which will surely be the third color jellybean (blue) as no other colour jellybean is left in her pocket
Answer - D

Avinashs87 · Answer

Shouldn't it be 9 actually...in case she picks G and B, then on 9th trial she can get a R.

Bunuel · Answer

For such cases when we should ensure, guarantee some outcome we always consider the worst case scenario. For 9 picks it's possible that Grace picks 8 R's and 1 G.

Check other  Worst Case Scenario Questions  from our  Special Questions Directory  to understand the concept better.

azhrhasan · Answer

13 (8+4+1) worst case

akadiyan · Answer

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?

A. 4
B. 8
C. 12
D. 13
E. 16

We need to maximize the worst case possibiilty for questions such as this

Total jellybeans = 16
Red =8
Green = 4
Blue = 4

Worst case can be first 8 draws are Red and next 4 are Green, even after 12 draws not all colors are represented. But in the 13th draw, the color has to be Blue. So minimum number of draws required is 13

Ans :   13

KhayatiK · Answer

Am i missing some crucial knowledge / piece of information here?! my approach was simply to subtract the minimum (smallest number possible (1)) number from each of the colored jellybeans since you can not get minimum than that. so red jelly beans (8-1 =7) +green (4-1=3)+blue (4-1=3) = 7+3+3 =13. but im looking at all the responses and they seem to be taking 1st 12 draws ; worse case scenario etc to approach the question.

Bunuel · Answer

Your method isn’t clear and doesn’t hold with other numbers.

For example, if a bag has 12 red, 9 green, and 5 blue marbles, your subtraction method gives 23.

Using the correct worst-case approach: you could draw all 12 red and all 9 green (21 marbles) and still not have a blue one. The next draw, the 22nd, must be blue, so the correct answer is 22.

This shows your shortcut fails. Study the worst-case method shown above and review the similar linked questions for more clarity and practice.

Grace has 16 jellybeans in her pocket. She has 8 red ones, 4

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Grace has 16 jellybeans in her pocket. She has 8 red ones, 4

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