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 by OmegaLearn

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

~ pi_is_3.14

https://www.youtube.com/watch?v=ZdZt9uzyMME

Solution

Video Solution- Quick, Easy Method

https://www.youtube.com/watch?v=dQ1EsX5JzoI

~Solution by Math Katana

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 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 3

We have 4 cases, if Tina chooses $1, 2, 3,$ or $4$ and always chooses numbers greater than the first number she chose.

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 sum 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 $\frac{7+6+5+4}{10 \cdot \binom{5}{2}}$ .

Case 2: Tina chooses $2$ .

In this case, Tina can choose $(2, 3), (2, 4),$ or $(2, 5).$

If Tina chooses $2$ and $3$ which sum to $5$ , Sergio only has $10-5=5$ 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 $\frac{5+4+3}{10 \cdot \binom{5}{2}}$ .

Case 3: Tina chooses $3$ .

In this case, Tina can choose $(3, 4),$ or $(3, 5).$

If Tina chooses $3$ and $4$ which sum to $7$ , Sergio only has $10-7=3$ 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 $\frac{3+2}{10 \cdot \binom{5}{2}}$ .

Case 4: Tina chooses $4$ .

In this case, Tina can only choose $(4,5).$

If Tina chooses $4$ and $5$ which sum to $9$ , Sergio only has $10-9=1$ choice.

Therefore, the probability of this is $\frac{1}{10 \cdot \binom{5}{2}}$ .

Once you add these probabilities up, you will have $\frac{(7+6+5+4)+(5+4+3)+(3+2)+(1)}{10 \cdot\binom{5}{2}} = \frac{40}{100} = \frac{2}{5}$ probability.

Thus our answer is $\frac{2}{5}$ .

~mathboy282

Solution 4

Assume Sergio chooses from ${2,3,\ldots,10}$ . The probability of Tina getting a sum of $6+x$ and a sum of $6-x$ , where $x \leq 4$ , 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 $\frac49 \cdot \frac{9}{10} + (0)\frac{1}{10} = \boxed{\textbf{(A)} \frac{2}{5}}.$

~zeric

Solution 5 (Brute Force)

List all the cases where $S \in [1, 10]$ and you get $\frac{0+0+0+1+2+4+6+8+9+10}{\binom{5}{2} \cdot 10} = \boxed{\textbf{(A)} \frac{2}{5}}$

~mathboy282

Solution 6 (Easy expected value solution)

The expected value of Tina is $\frac{1+2+3+4+5}{5}\cdot2=6$ , and there are 4 values greater than Tina (7, 8, 9, 10) out of 10. The probability is therefore $\frac{4}{10} = \boxed{\frac{2}{5}}$ .

2002 AMC 12A (Problems • Answer Key • Resources)
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2002 AMC 10A (Problems • Answer Key • Resources)
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2002 AMC 12A Problems/Problem 16

Contents

Problem

Video Solution by OmegaLearn

Solution

Video Solution- Quick, Easy Method

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5 (Brute Force)

Solution 6 (Easy expected value solution)

See Also