Collection of Maths Problems

Rings in soup II

Task number: 3540

• Variant

  What is the probability that the king will find the remaining (ruby) ring in his soup, under the assumption that he will first find the diamond ring there?

• Solution

  The event \(A\) where the king will find both rings has probability \(P(A)=\frac{1}{625}\).

  The probability of the event \(B\), where the king finds the diamong ring, is \(P(B)=\frac{25}{625}\).

  Because \(A\subseteq B\), we have \(P(A\cap B)=P(A)\).

  So now the probability \(P(A|B)=\frac{\frac{1}{625}}{\frac{25}{625}}=\frac{1}{25}\).

• Answer

  The probability that the king will also find the second (ruby) ring is \(\frac{1}{25}\).

• Variant

  What is the probability that the king will find both rings in his soup, under the assumption that he will find at least one?

• Solution

  The event \(A\) where the king will find both rings has probability \(P(A)=\frac{1}{625}\).

  The probability of the event \(B\) where the king will find at least one ring is \(P(B)=\frac{1}{625}+24\cdot\frac{2}{625}=\frac{49}{625}\).

  Because \(A\subseteq B\), we have \(P(A\cap B)=P(A)\).

  So the probabilty \(P(A|B)=\frac{\frac{1}{625}}{\frac{49}{625}}=\frac{1}{49}\).

  The probability is smaller compared to the first case, because we proceed from a broader assumption.

• Answer

  The probability that the king will also find a second ring is \(\frac{1}{49}\).

• Comment

  As a side note, this is not the same question as ‘The king has already found one of the rings; what is the probability that he will also find the second?‘ That has the same answer as the preceding variation, i.e. \(\frac{1}{25}\). At the moment when one of the rings has been found, it remains to find only one concrete ring, the remaining one. The fact that the question does not specify which ring must be found cannot influence the resulting probability.

  In other words, the situations ‘if the king were to find at least one of the two …‘ and ‘when the king will find one of the two …‘ are significantly different, even though they appear very similar. In the first case there are still two open possibilities as to which ring may be found.