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

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

Revision as of 16:19, 2 February 2024

Problem

Solution

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

See also

2024 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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