Brainsteem #10 Prize Mathematics Puzzle [Win 2 SBD and 1 SBD]
Remember that there are TWO prizes to aim for.
The experiment of paying out 40% and 10% has not really worked. I thought it would increase the overall prizes but this hasn't happened yet. What it has done is made it more time-consuming for me to find the earnings and then calculate the payouts. Therefore I am going back to fixed prizes and will increase them as earnings increase. One thing I may try is to use the Reward Distribution feature on chainBB and award the winner 40% of the next puzzle. You may comment if this is a good idea.
The Question
There are 21 balls in a bag. R of them are red, W of them are white, and B of them are blue. It is known that,
a) if we double the white balls, the probability of drawing one red ball is 1/28 less than the probability of drawing one red ball at the beginning,
and
b) if we remove all red balls, the probability of drawing one white ball is 1/18 more than the probability of drawing one white ball at the beginning.
Find R, W and B
The Prizes
To qualify you must upvote this post and write the answer in a comment.
To win First Prize you must include the answer and a valid method.
The First Prize will be 2 SBD.
The Second Prize will be 1 SBD, awarded to any other correct answer in the comments section, chosen at random after the post has expired.
Also remember that your upvote generates curation rewards for you and that 25% of all earnings are given to upvoters.
Prizes will be awarded after the 7 days have expired. Winners will be announced in the comments below after the draw.
The Prizes will increase with more participation and as the earnings increase.
The main thing is to enjoy the challenge and maybe learn something new.
Come and check out other challenges currently live!
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From the information provided we can obtain the following equations:-
R / (21+W) + 1/28 = R / 21
W / (W+B) = 1/18 + W / 21
R+W+B = 21
3 equations with 3 unknowns, we can obtain the answer by algebraic manipulation, which is R=3, W=7, B=11.
The probability of drawing a red ball is originally R/21, and clue "a" reduces that probability by 1/28 after adding balls. The greatest common factor (GCF) of 21 and 28 is 7, which tells me that W must be a multiple of 7. There can't be 21 white balls to begin with, as that would make the probability of drawing a red ball always 0. If W = 14, then R/21 becomes R/35.
No good. W = 7.
So trying again with W = 7 resulting in R/21 turning into R/28...
So far, R = 3 and W = 7 (and thus B = 11 as well) looks good. We just need to verify through clue "b". That clue gives us the equality W/21 + 1/18 = W/(21 - R). If we plug in R = 3 and W = 7 and simplify, both sides equal 7/18. Thus the solution has been verified. R = 3, W = 7, B = 11.
On the rewards issue, I think you should stick to the fixed prizes for now, as that seems to be the easiest way for you to handle it. The chainBB Reward Distribution feature should be tested first using a post not involving a puzzle. Maybe something like "comment with an interesting math fact you learned this week". Pick one or two of your favorite responses as winners, and see if the prize(s) distribute as you thought it should. If it works properly on the test post, then you can look into switching back to the percentage method again.
The answer is:
R = 3
W = 7
B=11
To solve it I just wrote the info as equations and by solving them I came up with the following two equations:
18xy = 441-21x
28xy=441+21y
(I changed R to x, W to y and B to z)
I solved one of the above to x and substitute in the other. I then came up with two solutions, one was not an integer which is not acceptable and so the result was the other one.
If you want a more detailed solution I can provide it.