Probabilistically Speaking, Romeo Needs To Try Harder

Let us interpret the problem from a different perspective. Let $n$ be the answer. Given $n$ hours that Romeo comes, what is the probability that, if Juliet chooses 2 hours, Romeo also chose both of those? To find this probability, we count the successful over the total. The successful is when she chooses both, which happens $\dbinom{n}{2}$ times. The total is the total number of possible choices of two hours, which happens $\dbinom{15}{2}$ times. Thus, $\dfrac{\dbinom{n}{2}}{\dbinom{15}{2}}\ge 0.7$ and we find $n\ge\boxed{13}$ .

Alternatively, we can WLOG Juliet chooses the first two hours, and proceed differently.

We label the $15$ possible times as $S=\{1,2, \cdots, 15\}$ . Suppose Juliet chooses times $i$ & $j$ with $1 \leq i <j \leq 15$ to stand at the balcony. Suppose Romeo will pass by at least $n$ times to ensure by $70\%$ seeing his love twice. Then, $i,j \in R$ , the set of times in which Romeo passes the balcony. $R \subset S \Rightarrow$ the probability of $i,j$ being in $R$ is at least $70 \%$ . The favourable case here is the consideration in which $i,j \in R$ & other $n-2$ times are chosen from rest $13$ times. That is,

$\frac{{{15-2} \choose {n-2}}}{{{15} \choose {n}}} \geq 70 \% \Rightarrow n(n-1) \geq 147$ . The minimum such $n$ is $\boxed{13}$ . So Romeo indeed has to strive to glimpse his love!

If there are $15$ possible times during the day that Juliet will be standing at the balcony, then there are ${15 \choose 2} = 105$ ways that Juliet may stand in the balcony one day. Then, not because of the multiple choice format, but we're looking for a $70\%$ , so we can conclude that the minimum times that Romeo will pass for a $70\%$ of chance will not be very far away from $15$ . The point is, that if Romeo passes $x$ times, then his chance of seeing Juliet twice a day is $\frac {{x \choose 2}}{{15 \choose 2}} = \frac {{x \choose 2}}{105}$ . Testing the values of $x$ that are close to $15$ (but smaller), we find that the minimum number of times for at least a $70\%$ chance is $\boxed {13}$ .

let romeo visits x times

total no. of ways of choosing 2 hours from 15 rs is C(15,2)

total no. of ways that she will choose 2 hrs from xhrs is C(x,2)

thus,

C(x,2)/C(15,2) > .7

=> x(x-1) > 0.7 15 14

=> x(x-1) > 147

minimum visits is 13

Let n be the number of times Romeo passes . Sample space=Number of ways in which Juliet can come out = 15C2. Number of suitable cases=Number of ways in which Romeo sees Juliet assuming she appears among the times Romeo passes= nC2. Probability=nC2 / 15C2 >= 0.7 The first such n arrives at n=13. So,answer is n=13