Difference between revisions of "2024 AIME I Problems/Problem 4"

Latest revision as of 18:26, 5 February 2024

Problem

Jen enters a lottery by picking $4$ distinct numbers from $S=\{1,2,3,\cdots,9,10\}.$ $4$ numbers are randomly chosen from $S.$ She wins a prize if at least two of her numbers were $2$ of the randomly chosen numbers, and wins the grand prize if all four of her numbers were the randomly chosen numbers. The probability of her winning the grand prize given that she won a prize is $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution 1

This is a conditional probability problem. Bayes' Theorem states that $P(A|B)=\dfrac{P(B|A)\cdot P(A)}{P(B)}$

in other words, the probability of $A$ given $B$ is equal to the probability of $B$ given $A$ times the probability of $A$ divided by the probability of $B$ . In our case, $A$ represents the probability of winning the grand prize, and $B$ represents the probability of winning a prize. Clearly, $P(B|A)=1$ , since by winning the grand prize you automatically win a prize. Thus, we want to find $\dfrac{P(A)}{P(B)}$ .

Let us calculate the probability of winning a prize. We do this through casework: how many of Jen's drawn numbers match the lottery's drawn numbers?

To win a prize, Jen must draw at least $2$ numbers identical to the lottery. Thus, our cases are drawing $2$ , $3$ , or $4$ numbers identical.

Let us first calculate the number of ways to draw exactly $2$ identical numbers to the lottery. Let Jen choose the numbers $a$ , $b$ , $c$ , and $d$ ; we have $\dbinom42$ ways to choose which $2$ of these $4$ numbers are identical to the lottery. We have now determined $2$ of the $4$ numbers drawn in the lottery; since the other $2$ numbers Jen chose can not be chosen by the lottery, the lottery now has $10-2-2=6$ numbers to choose the last $2$ numbers from. Thus, this case is $\dbinom62$ , so this case yields $\dbinom42\dbinom62=6\cdot15=90$ possibilities.

Next, let us calculate the number of ways to draw exactly $3$ identical numbers to the lottery. Again, let Jen choose $a$ , $b$ , $c$ , and $d$ . This time, we have $\dbinom43$ ways to choose the identical numbers and again $6$ numbers left for the lottery to choose from; however, since $3$ of the lottery's numbers have already been determined, the lottery only needs to choose $1$ more number, so this is $\dbinom61$ . This case yields $\dbinom43\dbinom61=4\cdot6=24$ .

Finally, let us calculate the number of ways to all $4$ numbers matching. There is actually just one way for this to happen.

In total, we have $90+24+1=115$ ways to win a prize. The lottery has $\dbinom{10}4=210$ possible combinations to draw, so the probability of winning a prize is $\dfrac{115}{210}$ . There is actually no need to simplify it or even evaluate $\dbinom{10}4$ or actually even know that it has to be $\dbinom{10}4$ ; it suffices to call it $a$ or some other variable, as it will cancel out later. However, let us just go through with this. The probability of winning a prize is $\dfrac{115}{210}$ . Note that the probability of winning a grand prize is just matching all $4$ numbers, which we already calculated to have $1$ possibility and thus have probability $\dfrac1{210}$ . Thus, our answer is $\dfrac{\frac1{210}}{\frac{115}{210}}=\dfrac1{115}$ . Therefore, our answer is $1+115=\boxed{116}$ .

~Technodoggo

Shortcut

One may also use complimentary counting as a shortcut to calculate the probability of winning a prize, in which the cases are that either one number is shared or no numbers are shared. There are $4 \cdot { {10 - 4} \choose {4-1}} = 4 \cdot 20 = 80$ ways to choose the former and ${{10-4} \choose 4} = 15$ ways for the latter. Therefore, there are $95$ ways to NOT choose a prize, so there are $210-95 = 115$ ways to choose a prize, and the answer follows.

- spectraldragon8

Solution 2

For getting all $4$ right, there is only $1$ way.

For getting $3$ right, there is $\dbinom43$ multiplied by $\dbinom61$ = $24$ ways.

For getting $2$ right, there is $\dbinom42$ multiplied by $\dbinom62$ = $90$ ways.

$\frac{1}{1+24+90}$ = $\frac{1}{115}$

Therefore, the answer is $1+115 = \boxed{116}$

~e___

Video Solution

https://youtu.be/pYU4RHRfe7M

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

@@ Line 1: / Line 1: @@
 ==Problem==
-Jen enters a lottery by picking <math>4</math> distinct numbers from <math>S=\{1,2,3,\cdots,9,10\}.</math> <math>4</math> numbers are randomly chosen from <math>S.</math> She wins a prize if at least two of her numbers were <math>2</math> of the randomly chosen numbers, and wins the grand prize if all four of her numbers were the randomly chosen numbers. The probability of her winning the grand prize given that she won a prize is <math>\frac{p}{q}</math> where <math>p,q</math> are relatively prime. What is the value of <math>p+q</math>?
+Jen enters a lottery by picking <math>4</math> distinct numbers from <math>S=\{1,2,3,\cdots,9,10\}.</math> <math>4</math> numbers are randomly chosen from <math>S.</math> She wins a prize if at least two of her numbers were <math>2</math> of the randomly chosen numbers, and wins the grand prize if all four of her numbers were the randomly chosen numbers. The probability of her winning the grand prize given that she won a prize is <math>\tfrac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
-==Solution==
+==Solution 1==
 This is a conditional probability problem. Bayes' Theorem states that
 <cmath>P(A|B)=\dfrac{P(B|A)\cdot P(A)}{P(B)}</cmath>
@@ Line 20: / Line 20: @@
 ~Technodoggo
+==Shortcut==
+One may also use complimentary counting as a shortcut to calculate the probability of winning a prize, in which the cases are that either one number is shared or no numbers are shared. There are <math>4 \cdot { {10 - 4} \choose {4-1}} = 4 \cdot 20 = 80</math> ways to choose the former and <math>{{10-4} \choose 4} = 15</math> ways for the latter. Therefore, there are <math>95</math> ways to NOT choose a prize, so there are <math>210-95 = 115</math> ways to choose a prize, and the answer follows.
+- [https://artofproblemsolving.com/wiki/index.php/User:Spectraldragon8 spectraldragon8]
 ==Solution 2==
@@ Line 25: / Line 30: @@
 For getting all <math>4</math> right, there is only <math>1</math> way.
-For getting <math>3</math> right, there is <math>\dbinom43</math> multiplied by <math>\dbinom61</math> = <math>24</math>.
+For getting <math>3</math> right, there is <math>\dbinom43</math> multiplied by <math>\dbinom61</math> = <math>24</math> ways.
+For getting <math>2</math> right, there is <math>\dbinom42</math> multiplied by <math>\dbinom62</math> = <math>90</math> ways.
+<math>\frac{1}{1+24+90}</math>  = <math>\frac{1}{115}</math>
+Therefore, the answer is <math>1+115 = \boxed{116}</math>
+~e___
+==Video Solution==
+https://youtu.be/pYU4RHRfe7M
-For getting <math>2</math> right, there is <math>\dbinom42</math> multiplied by <math>\dbinom62</math> = <math>90</math>.
+~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
-<math>1/(1+24+90)</math> = \dfrac1{115}<math>.
-Therefore, the answer is </math>1+115=<math> </math>\boxed{116}$
 ==See also==

2024 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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