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

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<math>\text{(A)}\ 2/5 \qquad \text{(B)}\ 9/20 \qquad \text{(C)}\ 1/2 \qquad \text{(D)}\ 11/20 \qquad \text{(E)}\ 24/25</math>
 
<math>\text{(A)}\ 2/5 \qquad \text{(B)}\ 9/20 \qquad \text{(C)}\ 1/2 \qquad \text{(D)}\ 11/20 \qquad \text{(E)}\ 24/25</math>
  
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== Video Solution ==
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== Video Solution by OmegaLearn ==
 
https://youtu.be/8WrdYLw9_ns?t=381
 
https://youtu.be/8WrdYLw9_ns?t=381
  
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https://www.youtube.com/watch?v=ZdZt9uzyMME
 
https://www.youtube.com/watch?v=ZdZt9uzyMME
  
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==Solution==
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=Solution=
 +
 
 +
==Video Solution- Quick, Easy Method==
 +
https://www.youtube.com/watch?v=dQ1EsX5JzoI
 +
 
 +
~Solution by Math Katana
  
 
=== Solution 1 ===
 
=== Solution 1 ===
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==Solution 3==
 
==Solution 3==
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We have 5 cases, if Tina choose <math>1, 2, 3, 4,</math> or <math>5.</math>
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We have 4 cases, if Tina chooses <math>1, 2, 3,</math> or <math>4</math> and always chooses numbers greater than the first number she chose.
  
 
The number of ways of choosing 2 numbers from <math>5</math> are <math>\binom{5}{2}</math>.
 
The number of ways of choosing 2 numbers from <math>5</math> are <math>\binom{5}{2}</math>.
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Therefore, the probability of this is <math>\frac{7+6+5+4}{10 \cdot \binom{5}{2}}</math>.
 
Therefore, the probability of this is <math>\frac{7+6+5+4}{10 \cdot \binom{5}{2}}</math>.
 
--------------------------
 
--------------------------
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If you do this over and over again you will see that you have <math>\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}</math> probability.
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Case 2: Tina chooses <math>2</math>.
 +
 
 +
In this case, Tina can choose <math>(2, 3), (2, 4),</math> or <math>(2, 5).</math>
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 +
If Tina chooses <math>2</math> and <math>3</math> which sum to <math>5</math>, Sergio only has <math>10-5=5</math> choices.
 +
 
 +
Since the sum of the combined numbers increases by <math>1</math> every time for this specific case, Sergio has <math>1</math> less choice every time.
 +
 
 +
Therefore, the probability of this is <math>\frac{5+4+3}{10 \cdot \binom{5}{2}}</math>.
 +
--------------------------
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Case 3: Tina chooses <math>3</math>.
 +
 
 +
In this case, Tina can choose <math>(3, 4),</math> or <math>(3, 5).</math>
 +
 
 +
If Tina chooses <math>3</math> and <math>4</math> which sum to <math>7</math>, Sergio only has <math>10-7=3</math> choices.
 +
 
 +
Since the sum of the combined numbers increases by <math>1</math> every time for this specific case, Sergio has <math>1</math> less choice every time.
 +
 
 +
Therefore, the probability of this is <math>\frac{3+2}{10 \cdot \binom{5}{2}}</math>.
 +
--------------------------
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Case 4: Tina chooses <math>4</math>.
  
−
But since we overcounted by 2 (e.g. <math>(1, 2)</math> and <math>(2, 1)</math>) we need to divide by <math>2.</math>
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In this case, Tina can only choose <math>(4,5).</math>
  
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Thus our answer is <math>\frac{4}{5} \div 2 = \boxed{\textbf{(A)} \frac{2}{5}}.</math>
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If Tina chooses <math>4</math> and <math>5</math> which sum to <math>9</math>, Sergio only has <math>10-9=1</math> choice.
  
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~mathboy282
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Therefore, the probability of this is <math>\frac{1}{10 \cdot \binom{5}{2}}</math>.
 +
--------------------------
 +
 
 +
Once you add these probabilities up, you will have <math>\frac{(7+6+5+4)+(5+4+3)+(3+2)+(1)}{10 \cdot\binom{5}{2}} = \frac{40}{100} = \frac{2}{5}</math> probability.
  
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Note: I will add in all the cases soon, kind of busy today so yea.
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Thus our answer is <math>\frac{2}{5}</math>.
  
 +
~mathboy282
 
=== Solution 4 ===
 
=== Solution 4 ===
  
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~mathboy282
 
~mathboy282
  
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== Solution 6 ==
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== Solution 6 (Easy expected value solution) ==
  
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The expected value of Tina is <math>\frac{1+2+3+4+5}{5}\cdot2=6</math>, and there are 4 values (7, 8, 9, 10) that are more than 6 out of 10. The probability is therefore <math>\frac{4}{10} = \boxed{\frac{2}{5}}</math>.
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The expected value of Tina is <math>\frac{1+2+3+4+5}{5}\cdot2=6</math>, and there are 4 values greater than Tina (7, 8, 9, 10) out of 10. The probability is therefore <math>\frac{4}{10} = \boxed{\frac{2}{5}}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 21:43, 16 October 2024

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 $\frac{4}{10} = \boxed{\frac{2}{5}}$.

See Also

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

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