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

(New page: {{duplicate|2002 AMC 12A #16 and 2002 AMC 10A #24}} ==Problem== Tina randomly selects two distinct numbers from the set {1, 2, 3, 4, 5...)
 
(→‎Solution 3)
 
(44 intermediate revisions by 18 users not shown)
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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) \\
 +
&= \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>.
 +
--------------------------
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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>.
 +
 
 +
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 ===
 +
 
 +
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 4: This once again happens in only one way (1,3). Sergio can choose a number from 5 to 10, so 6 ways here.
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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>.  
  
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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.
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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 6: Two ways here (1,5) and (2,4). Sergio can choose a number from 7 to 10, so 2*4=8 here.
+
~zeric
  
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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.
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===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 8: Only one way possible (3,5). Sergio chooses 9 or 10, so 2 ways here.
+
~mathboy282
  
−
Tina gets a sum of 9: Only one way (4,5). Sergio must choose 10, so 1 way.
+
== Solution 6 (Easy expected value solution) ==
  
−
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>.
+
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==
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{{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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