Difference between revisions of "2002 AMC 12A Problems/Problem 16"

Revision as of 18:33, 8 August 2022

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 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 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 4

Assume Sergio chooses from ${2,3,\ldots,10}$ . The probability 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

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

@@ Line 83: / Line 83: @@
 === Solution 4 ===
-Assume Sergio chooses from <math>{2,3,\ldots,10}</math>. The probably of Tina getting a sum of <math>6+x</math> and <math>6-x</math> (<math>x \leq 4</math>) are equal due to symmetry. The probability of Sergio choosing numbers higher/lower than <math>6+x</math> is equal to him choosing numbers lower/higher than <math>6-x</math>. 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.
+Assume Sergio chooses from <math>{2,3,\ldots,10}</math>. The probability of Tina getting a sum of <math>6+x</math> and <math>6-x</math> (<math>x \leq 4</math>) are equal due to symmetry. The probability of Sergio choosing numbers higher/lower than <math>6+x</math> is equal to him choosing numbers lower/higher than <math>6-x</math>. 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 <math>1/9</math>, so the probability of Sergio getting a higher number is <math>\frac{(9-1)/2}{9} = \frac49</math>.

2002 AMC 12A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

Page

Toolbox

Search