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Ques: A covered basket of flowers has some lilies and roses. In search of rose, Sweety and Shweta alternately pick up a flower from the basket but put it back if it is not a rose. Sweety is 3 times more likely to be the first one to pick a rose. If sweety begin this 'rose hunt' and if there are 60 lilies in the basket, find the number of roses in the basket.

Let the roses be x and lilies be y. Since they find all of the x roses finally $$ x!/(x+y)! + y/(x+y) * x!/(x+y)! \cdots = 1 $$ So to be straight forward, I do not have any good idea how to deal with this.

CodeBlooded
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2 Answers2

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Assume that the probability to pick a rose is $r$. Then - if $p$ denotes the probability that Sweety (who starts picking) will be the first to pick a rose - we find: $$p=r+(1-r)^2p\tag1$$

Here $r$ is the probability that her first pick is succesful and $(1-r)^2$ is the probability that both are unsuccesful at their first pick so that the process starts over with again a probability $p$ that Sweety will be the first one with success.

Next to that we also have:$$p=3(1-p)$$or equivalently:$$p=\frac34$$and this makes it possible to solve $(1)$.

One solution is: $r=\frac23$ telling us that next to $60$ lillies there are $120$ roses.

Another solution is $r=0$ telling us that there are no roses at all. Then Sweety and Shweta are still picking at the moment, both having probability $0$ to succeed once. Observe that $0=3\cdot0$ so that indeed the probability of Sweety to be the first is $3$ times the probability of Shweta to be the first.

Actually this second solution could be excluded on base of the info that the basket contains lillies and roses (so $r>0$).

drhab
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  • so the fact that Sweety is more likely to get a rose is just based on the fact that she's the first to start picking, right? Though I don't think it's specified in the question. – harry Apr 16 '21 at 09:00
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    @HarryHolmes Yes, it is purely based on the fact that Sweety starts. In the question we find the words: "If sweety begins this 'rose hunt' and if..." So in that sense it is specified in the question. Also if nothing is said about it then on base of the fact that Sweety has larger chances and they pick alternatively we can conclude that Sweety must be the one who starts. – drhab Apr 16 '21 at 09:06
  • Thanks, great answer with the algebraic probabilities. – harry Apr 16 '21 at 09:07
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Let $x$ be the number of roses in the basket.

It's given that the first person to choose is $3$ times more likely to get the first rose than the second person, hence the first person gets the first rose with probability $\frac{3}{4}$.

With probability ${\large{\frac{x}{x+60}}}$, Sweety wins on the first try.

With probability ${\large{\frac{60}{x+60}}}$, Sweety does not win on the first try. In that case, Shweta is effectively the first player in a new game. Hence, starting at that point, Shweta has probability $\frac{3}{4}$ to win, and Sweety has probability $\frac{1}{4}$ to win.

This yields the equation $$\frac{3}{4}=\frac{x}{x+60}+\left(\frac{60}{x+60}\right)\left(\frac{1}{4}\right)$$

Solving for $x$ yields $x = 120$.

quasi
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  • in the final equation, the first term signifies the probability that Sweety wins on the first try, and the second term shows the combination of the probability of her losing the first try and winning the second, right? And just analysing these two rounds are enough, because from there on, the cycle repeats. Have I got it right? – harry Apr 16 '21 at 09:27
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    @Harry Holmes: Partially right. Yes, the first term $$ \frac{x}{x+60} $$ is the probability that Sweeny wins on the first try, but the second term $$ \left(\frac{60}{x+60}\right)\left(\frac{1}{4}\right) $$ is the probability that Sweety fails on the first try and then, with Shweta now effectively going first, Shweta is not the first to get a rose. – quasi Apr 16 '21 at 19:10
  • Right, got it, thanks. – harry Apr 17 '21 at 00:35