M40-35

Question

A jar contains only green and red candies. If the ratio of green candies to red candies in the jar is 8 to 3, what is the number of red candies in the jar?

(1) The minimum number of candies one must pick from the jar to ensure getting at least one candy of each color is 25.
 
(2) The probability of randomly picking a red candy from the jar is  \frac{3}{11} .

Bunuel · Answer

Official Solution:

A jar contains only green and red candies. If the ratio of green candies to red candies in the jar is 8 to 3, what is the number of red candies in the jar? 
 
(1) The minimum number of candies one must pick from the jar to ensure getting at least one candy of each color is 25.
 
To guarantee that we get at least one candy of each color, we should first pick all candies of the larger count, and then the next one will be of the other color. So, the minimum being 25 indicates that there are 24 green candies in the jar. Since the ratio of green candies to red candies in the jar is 8 to 3, then there must be 9 red candies. Sufficient.
 
(2) The probability of randomly picking a red candy from the jar is  \frac{3}{11} .
 
This statement merely restates the given information in the question stem: since the ratio of green candies to red candies in the jar is 8 to 3, the probability of picking a red candy from the jar is indeed 3/11. Not sufficient.

Answer: A

Rebaz · Answer

This question is in my opinion a great and a tricky question.

I made the assumption that to ensure getting at least one candy of each color for it to be 25, then we must have 24 candies of each colour.

But if you read the question carefully and see that the ratio of green candies to red candies in the jar is 8 to 3, then you know that we have  more Green candies   than  Red Candies , and for this to be the case, we can not have 24 candies of each colour.

The only option left is 24 green Candies and 9 Red candies and we must pick all the green candies (24) first and then a red candy, which is what statement 1 is exactly saying.

Once again, a very Tricky and Great Question!

Thanks for posting it!

Chloee · Answer

I have a question about this status; (1) The minimum number of candies one must pick from the jar to ensure getting at least one candy of each color is 25.

I understood that there should be 24 same candies, but I thought it could be red like this-
G:R=64:24

If there are 24 red candies, wouldn't it be possible to draw one green candy after drawing all the red candies? 64:24 is also a ratio of 3/11.

Please help me! Thanks.

Bunuel · Answer

No. Having 64 green candies and 24 red candies won't satisfy the first statement. In this case the minimum number of candies one must pick from the jar to ensure getting at least one candy of each color would be 64 + 1 = 65.

To understand the concept better, check other  Worst Case Scenario Questions  from our  Special Questions Directory  for additional practice.

Chloee · Answer

Now I fully understood; If there are 64 green candies, there is no guarantee that the 25th candy drawn will be a red candy. Because even after drawing 24 of the 64 green candies, there are still 40 green candies left. I was a fool lol

Thanks a lot Bunuel!

Thanks a lot Bunuel!

Abhishake03 · Answer

I think that equation (l) still doesn't solve the equation as the question remains if the number of red candies are 24 or 9. The statement is saying that minimum number of candies one must pick to ensure both type of candies are picked is 25. Then 24 can either be red candies or green candies. If 24 are green candies than red candies would be 9 and vice versa.

Bunuel · Answer

No, the color with more candies must be the one you can pick 24 times without getting both colors. Since the green-to-red ratio is 8:3, green candies are more numerous. Thus, the 24 candies must be green, not red. Also, if Red = 24 and Green = 64, the minimum number of picks needed to ensure both colors would be 64 + 1 = 65, not 25. So Red cannot be 24. Please review carefully.

JayathS · Answer

I like the solution - it’s helpful.

Saun2511 · Answer

Bunuel while rare but it is not impossible to get 24 red candies first and the 25th being green even if green is in majority. So from first statement, we know that ONE of the colored candies is 24 in number in the bowl. I dont think we can for sure say that its only green because there are more of those.

Bunuel · Answer

You’re mixing “possible” with “guaranteed.”

“Minimum number to ensure both colors” is computed in the worst-case draw: you could pull all candies of the more numerous color first, then one more pick forces the other color. With ratio 8:3, green is always the majority. So the worst-case run is all greens. If the minimum is 25, the majority count must be 24. Hence greens = 24, and by 8:3, one part is 3, so reds = 9.

Your “24 reds first” scenario is irrelevant to the guarantee and also incompatible with the data: if red = 24, then by 8:3 green = 64, and the minimum to ensure both would be 64 + 1 = 65, not 25.

To understand the concept better, check other  Worst Case Scenario Questions  from our  Special Questions Directory  for additional practice.

shumalik · Answer

I did not quite understand the solution. why do we say that we start with larger count of candies to guarantee each candy being picked. I approached the solution in a similar way that there would be 24 candies of any colour but how do we ensure that, this number is of green(larger) candies?

Bunuel · Answer

I tried elaborating the solution in a few posts above, so please go through them. I also included links to similar questions that can help you understand the concept and practice it. Check those questions as well for additional context.

Vivek1707 · Answer

This is a great question that’s helpful for learning.

M40-35

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