2019 AIME I Problems/Problem 2

Problem 2

Jenn randomly chooses a number $J$ from $1, 2, 3,\ldots, 19, 20$. Bela then randomly chooses a number $B$ from $1, 2, 3,\ldots, 19, 20$ distinct from $J$. The value of $B - J$ is at least $2$ with a probability that can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.


By symmetry, the desired probability is equal to the probability that $J - B$ is at most $-2$, which is $\frac{1-P}{2}$ where $P$ is the probability that $B$ and $J$ differ by $1$ (no zero, because the two numbers are distinct). There are $20 * 19 = 380$ total possible combinations of $B$ and $J$, and $1 + 18 * 2 + 1 = 38$ ones that form $P$, so $P = \frac{38}{380} = \frac{1}{10}$. Therefore the answer is $\frac{9}{20} \rightarrow \boxed{029}$.

Solution 2

This problem is essentially asking how many ways there are to choose $2$ distinct elements from a $20$ element set such that no $2$ elements are adjacent. Using the well-known formula $\dbinom{n-k+1}{k}$, there are $\dbinom{20-2+1}{2} = \dbinom{19}{2} = 171$ ways. Dividing $171$ by $380$, our desired probability is $\frac{171}{380} = \frac{9}{20}$. Thus, our answer is $9+20=\boxed{029}$. -Fidgetboss_4000

Solution 3

Create a grid using graph paper, with $20$ columns for the values of $J$ from $1$ to $20$ and $20$ rows for the values of $B$ from $1$ to $20$. Since $B$ cannot equal $J$, we cross out the diagonal line from the first column of the first row to the twentieth column of the last row. Now, since $B - J$ must be at least $2$, we can mark the line where $B - J = 2$. Now we sum the number of squares that are on this line and below it. We get $171$. Then we find the number of total squares, which is $400 - 20 = 380$. Finally, we take the ratio $\frac{171}{380}$, which simplifies to $\frac{9}{20}$. Our answer is $9+20=\boxed{029}$.

Solution 4

We can see that if $B$ chooses $20$, $J$ can choose from $1$ through $18$ such that $B-J\geq 2$. If $B$ chooses $19$, $J$ has choices $1$~$17$. By continuing this pattern, $B$ will choose $3$ and $J$ will have $1$ option. Summing up the total, we get $18+17+\cdots+1$ as the total number of solutions. The total amount of choices is $20\times19$ (B and J must choose different numbers), so the probability is $\frac{18*19/2}{20*19}=\frac{9}{20}$. Therefore, the answer is $9+20=\boxed{029}$


Solution 5

Similar to solution 4, we can go through the possible values of $J$ to find all the values of $B$ that makes $B-J\geq 2$. If $J$ chooses $1$, then $B$ can choose anything from $3$ to $20$. If $J$ chooses $2$, then $B$ can choose anything from $4$ to $20$. By continuing this pattern, we can see that there is $18+17+\cdots+1$ possible solutions. The amount of solutions is, therefore, $\frac{18*19/2}=171$. Now, because $B$ and $J$ must be different, we have $20\times19=380$ possible choices, so the probability is $\frac{171}{380}=\frac{9}{20}$. Therefore, the final answer is $9+20=\boxed{029}$


Video Solution


Video Solution 2



Video Solution 3


See Also

2019 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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