2002 AMC 12A Problems/Problem 16

Revision as of 00:00, 14 March 2021 by Bronzetruck2016 (talk | contribs) (Solution 2)
The following problem is from both the 2002 AMC 12A #16 and 2002 AMC 10A #24, so both problems redirect to this page.


Problem

Tina randomly selects two distinct numbers from the set $\{ 1, 2, 3, 4, 5 \}$, and Sergio randomly selects a number from the set $\{ 1, 2, ..., 10 \}$. What is the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina?

$\text{(A)}\ 2/5 \qquad \text{(B)}\ 9/20 \qquad \text{(C)}\ 1/2 \qquad \text{(D)}\ 11/20 \qquad \text{(E)}\ 24/25$

Video Solution

https://youtu.be/8WrdYLw9_ns?t=381

~ pi_is_3.14

Solution

Solution 1

This is not too bad using casework.

Tina gets a sum of 3: This happens in only one way $(1,2)$ and Sergio can choose a number from 4 to 10, inclusive. There are 7 ways that Sergio gets a desirable number here.

Tina gets a sum of 4: This once again happens in only one way $(1,3)$. Sergio can choose a number from 5 to 10, so 6 ways here.

Tina gets a sum of 5: This can happen in two ways $(1,4)$ and $(2,3)$. Sergio can choose a number from 6 to 10, so $2\cdot5=10$ ways here.

Tina gets a sum of 6: Two ways here $(1,5)$ and $(2,4)$. Sergio can choose a number from 7 to 10, so $2\cdot4=8$ here.

Tina gets a sum of 7: Two ways here $(2,5)$ and $(3,4)$. Sergio can choose from 8 to 10, so $2\cdot3=6$ ways here.

Tina gets a sum of 8: Only one way possible $(3,5$). Sergio chooses 9 or 10, so 2 ways here.

Tina gets a sum of 9: Only one way $(4,5)$. Sergio must choose 10, so 1 way.

In all, there are $7+6+10+8+6+2+1=40$ ways. Tina chooses two distinct numbers in $\binom{5}{2}=10$ ways while Sergio chooses a number in $10$ ways, so there are $10\cdot 10=100$ ways in all. Since $\frac{40}{100}=\frac{2}{5}$, our answer is $\boxed{\text{(A)}\frac{2}{5}}$.

Solution 2

We want to find the average of the smallest possible chance of Sergio winning and the largest possible chance of Sergio winning. This is because the probability decreases linearly. The largest possibility of Sergio winning if Tina chooses a 1 and a 1. The chances of Sergio winning is then $\frac{8}{10} = \frac{4}{5}$ . The smallest possibility of Sergio winning is if Tina chooses a 5 and a 5. The chances of Sergio winning then is $\frac{0}{10} = 0$. The average of $\frac{4}{5}$ and $0$ is $\boxed{\text{(A)}\frac{2}{5}}$.

-Minor edits by bronzetruck2016

Solution 3

We invoke some symmetry. Let $T$ denote Tina's sum, and let $S$ denote Sergio's number. Observe that, for $i = 2, 3, \ldots, 10$, $\text{Pr}(T=i) = \text{Pr}(T=12-i)$.

If Tina's sum is $i$, then the probability that Sergio's number is larger than Tina's sum is $\frac{10-i}{10}$. Thus, the probability $P$ is

\[P = \text{Pr}(S>T) = \sum_{i=2}^{10} \text{Pr}(T=i) \times \frac{10-i}{10}\]

Using the symmetry observation, we can also write the above sum as \[P = \sum_{i=2}^{10} \text{Pr}(T=12-i) \times \frac{10-i}{10} = \sum_{i=2}^{10} \text{Pr}(T=i) \times \frac{i-2}{10}\] where the last equality follows as we reversed the indices of the sum (by replacing $12-i$ with $i$). Thus, adding the two equivalent expressions for $P$, we have

\begin{align*} 2P &= \sum_{i=2}^{10} \text{Pr}(T=i) \times \left(\frac{10-i}{10} + \frac{i-2}{10}\right) \\ &= \sum_{i=2}^{10} \text{Pr}(T=i) \times \frac{4}{5} \\ &= \frac{4}{5} \sum_{i=2}^{10} \text{Pr}(T=i) \\ &= \frac{4}{5} \end{align*}

Since this represents twice the desired probability, the answer is $P = \boxed{\textbf{(A)} \frac{2}{5}}$. -scrabbler94

Solution 4

We have 5 cases, if Tina choose $1, 2, 3, 4,$ or $5.$

The number of ways of choosing 2 numbers from $5$ are $\binom{5}{2}$.


Case 1: Tina chooses $1$.

In this case, since the numbers are distinct, Tina can choose $(1, 2), (1, 3), (1, 4),$ or $(1, 5).$

If Tina chooses $1$ and $2$ which some to $3$, Sergio only has $10-3=7$ choices.

Since the sum of the combined numbers increases by $1$ every time for this specific case, Sergio has $1$ less choice every time.

Therefore, the probability of this is $\frac{7+6+5+4}{10 \cdot \binom{5}{2}}$.


If you do this over and over again you will see that you have $\frac{(7+6+5+4)+(7+5+4+3)+(6+5+3+2)+(5+4+3+1)+(4+3+2+1)}{10 \cdot \binom{5}{2}} = \frac{80}{100} = \frac{4}{5}$ probability.

But since we overcounted by 2 (e.g. $(1, 2)$ and $(2, 1)$) we need to divide by $2.$

Thus our answer is $\frac{4}{5} \div 2 = \boxed{\textbf{(A)} \frac{2}{5}}.$

~mathboy282

Note: I will add in all the cases soon, kind of busy today so yea.

Solution 5

Assume Sergio chooses from ${2,3,\ldots,10}$. The probably of Tina getting a sum of $6+x$ and $6-x$ ($x \leq 4$) are equal due to symmetry. The probability of Sergio choosing numbers higher/lower than $6+x$ is equal to him choosing numbers lower/higher than $6-x$. Therefore over all of Tina's sums, the probability of Sergio choosing a number higher is equal to the probability of choosing a number lower.

The probability that they get the same value is $1/9$, so the probability of Sergio getting a higher number is $\frac{(9-1)/2}{9} = \frac49$.

Sergio never wins when choosing $1$ so the probability is $\frac49 \frac{9}{10} + (0)\frac{1}{10} = \boxed{\textbf{(A)} \frac{2}{5}}.$

~zeric

See Also

2002 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
2002 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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