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Difference between revisions of "2002 AMC 12A Problems/Problem 16"
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==Problem==
 
==Problem==
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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?  
+
Tina randomly selects two distinct numbers from the set <math>\{ 1, 2, 3, 4, 5 \}</math>, and Sergio randomly selects a number from the set <math>\{ 1, 2, ..., 10 \}</math>. What is the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina?  
  
 
<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>
  
−
==Solution==
+
== 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.  
 
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 3: This happens in only one way <math>(1,2)</math> 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 <math>(1,3)</math>. 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 <math>(1,4)</math> and <math>(2,3)</math>. Sergio can choose a number from 6 to 10, so <math>2\cdot5=10</math> ways here.
 +
 
 +
Tina gets a sum of 6: Two ways here <math>(1,5)</math> and <math>(2,4)</math>. Sergio can choose a number from 7 to 10, so <math>2\cdot4=8</math> here.
 +
 
 +
Tina gets a sum of 7: Two ways here <math>(2,5)</math> and <math>(3,4)</math>. Sergio can choose from 8 to 10, so <math>2\cdot3=6</math> ways here.
 +
 
 +
Tina gets a sum of 8: Only one way possible <math>(3,5</math>). Sergio chooses 9 or 10, so 2 ways here.
 +
 
 +
Tina gets a sum of 9: Only one way <math>(4,5)</math>. Sergio must choose 10, so 1 way.
 +
 
 +
In all, there are <math>7+6+10+8+6+2+1=40</math> ways. Tina chooses two distinct numbers in <math>\binom{5}{2}=10</math> ways while Sergio chooses a number in <math>10</math> ways, so there are <math>10\cdot 10=100</math> ways in all. Since <math>\frac{40}{100}=\frac{2}{5}</math>, our answer is <math>\boxed{\text{(A)}\frac{2}{5}}</math>.
 +
 
 +
=== Solution 2 ===
 +
We invoke some symmetry. Let <math>T</math> denote Tina's sum, and let <math>S</math> denote Sergio's number. Observe that, for <math>i = 2, 3, \ldots, 10</math>, <math>\text{Pr}(T=i) = \text{Pr}(T=12-i)</math>.
 +
 
 +
If Tina's sum is <math>i</math>, then the probability that Sergio's number is larger than Tina's sum is <math>\frac{10-i}{10}</math>. Thus, the probability <math>P</math> is
 +
 
 +
<cmath>P = \text{Pr}(S>T) = \sum_{i=2}^{10} \text{Pr}(T=i) \times \frac{10-i}{10}</cmath>
 +
 
 +
Using the symmetry observation, we can also write the above sum as
 +
<cmath> 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}</cmath>
 +
where the last equality follows as we reversed the indices of the sum (by replacing <math>12-i</math> with <math>i</math>). Thus, adding the two equivalent expressions for <math>P</math>, we have
 +
 
 +
<cmath>
 +
\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) \\
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&= \frac{4}{5}
 +
\end{align*}
 +
</cmath>
 +
 
 +
Since this represents twice the desired probability, the answer is <math>P = \boxed{\textbf{(A)} \frac{2}{5}}</math>. -scrabbler94
 +
 
 +
==Solution 3==
 +
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>.
 +
--------------------------
 +
Case 1: Tina chooses <math>1</math>.
 +
 
 +
In this case, since the numbers are distinct, Tina can choose <math>(1, 2), (1, 3), (1, 4),</math> or <math>(1, 5).</math>
 +
 
 +
If Tina chooses <math>1</math> and <math>2</math> which sum to <math>3</math>, Sergio only has <math>10-3=7</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{7+6+5+4}{10 \cdot \binom{5}{2}}</math>.
 +
--------------------------
 +
Case 2: Tina chooses <math>2</math>.
 +
 
 +
In this case, Tina can choose <math>(2, 3), (2, 4),</math> or <math>(2, 5).</math>
 +
 
 +
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>.
 +
--------------------------
 +
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>.
 +
--------------------------
 +
Case 4: Tina chooses <math>4</math>.
 +
 
 +
In this case, Tina can only choose <math>(4,5).</math>
 +
 
 +
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.
 +
 
 +
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.
 +
 
 +
Thus our answer is <math>\frac{2}{5}</math>.
 +
 
 +
~mathboy282
 +
=== Solution 4 ===
  
−
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.
+
Assume Sergio chooses from <math>{2,3,\ldots,10}</math>. The probability of Tina getting a sum of <math>6+x</math> and a sum of <math>6-x</math>, where <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.  
  
−
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*5=10 ways here.
+
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>.  
  
−
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*4=8 here.
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Sergio never wins when choosing <math>1</math> so the probability is <math>\frac49 \cdot \frac{9}{10} + (0)\frac{1}{10} = \boxed{\textbf{(A)} \frac{2}{5}}.</math>
  
−
Tina gets a sum of 7: Two ways here (2,5) and (3,4). Sergio can choose from 8 to 10, so 2*3=6 ways here.
+
~zeric
  
−
Tina gets a sum of 8: Only one way possible (3,5). Sergio chooses 9 or 10, so 2 ways here.
+
===Solution 5 (Brute Force)===
 +
List all the cases where <math>S \in [1, 10]</math> and you get <math>\frac{0+0+0+1+2+4+6+8+9+10}{\binom{5}{2} \cdot 10} = \boxed{\textbf{(A)} \frac{2}{5}}</math>
  
−
Tina gets a sum of 9: Only one way (4,5). Sergio must choose 10, so 1 way.
+
~mathboy282
  
−
In all, there are <math>7+6+10+8+6+2+1=40</math> ways. Tina chooses two distinct numbers in <math>\binom{5}{2}=10</math> ways while Sergio chooses a number in <math>10</math> ways, so there are <math>10\cdot 10=100</math> ways in all. Since <math>\frac{40}{100}=\frac{2}{5}</math>, our answer is <math>\boxed{\text{(A)}\ 2/5}</math>.
+
== Solution 6 (Easy expected value solution) ==
  
−
An alternative way is 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 2. The chances of Sergio winning is then <math> \frac{7}{10}</math> . The smallest possibility of Sergio winning is if Tina chooses a 4 and a 5. The chances of Sergio winning then is <math> \frac{1}{10}. The average of </math> \frac{7}{10} <math> and </math> \frac{1}{10} <math> is </math>\boxed{frac{2}{5}}$.
+
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==
Line 32: Line 135:
 
{{AMC10 box|year=2002|ab=A|num-b=23|num-a=25}}
 
{{AMC10 box|year=2002|ab=A|num-b=23|num-a=25}}
  
 +
[[Category:Introductory Probability Problems]]
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
 +
{{MAA Notice}}

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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