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Combinatorics Difficulty 6.6 National olympiad Find the answer

Ali and Veli goes to hunting. The probability that each will successfully hit a duck is 1/21/2 on any given shot. During the hunt, Ali shoots 1212 times, and Veli shoots 1313 times. What is the probability that Veli hits more ducks than Ali?

Pick one

Solution

1. Define the random variables:
Let X X be the number of ducks hit by Ali, and Y Y be the number of ducks hit by Veli. Both X X and Y Y follow a binomial distribution:
XBinomial(12,12) X \sim \text{Binomial}(12, \frac{1}{2})
YBinomial(13,12) Y \sim \text{Binomial}(13, \frac{1}{2})

2. Calculate the expected values and variances:
The expected value and variance for a binomial distribution Binomial(n,p) \text{Binomial}(n, p) are given by:
E[X]=npandVar(X)=np(1p) E[X] = np \quad \text{and} \quad \text{Var}(X) = np(1-p)
For Ali:
E[X]=1212=6andVar(X)=121212=3 E[X] = 12 \cdot \frac{1}{2} = 6 \quad \text{and} \quad \text{Var}(X) = 12 \cdot \frac{1}{2} \cdot \frac{1}{2} = 3
For Veli:
E[Y]=1312=6.5andVar(Y)=131212=3.25 E[Y] = 13 \cdot \frac{1}{2} = 6.5 \quad \text{and} \quad \text{Var}(Y) = 13 \cdot \frac{1}{2} \cdot \frac{1}{2} = 3.25

3. Standardize the variables:
To find the probability that Y>X Y > X , we can standardize X X and Y Y to approximate them using the normal distribution (Central Limit Theorem):
ZX=X63andZY=Y6.53.25 Z_X = \frac{X - 6}{\sqrt{3}} \quad \text{and} \quad Z_Y = \frac{Y - 6.5}{\sqrt{3.25}}

4. Calculate the difference of the standardized variables:
Let Z=ZYZX Z = Z_Y - Z_X . Since ZX Z_X and ZY Z_Y are approximately standard normal variables, Z Z will also be approximately normal with:
E[Z]=E[ZY]E[ZX]=0 E[Z] = E[Z_Y] - E[Z_X] = 0
Var(Z)=Var(ZY)+Var(ZX)=1+1=2 \text{Var}(Z) = \text{Var}(Z_Y) + \text{Var}(Z_X) = 1 + 1 = 2
Thus, ZN(0,2) Z \sim N(0, 2) .

5. Find the probability:
We need to find P(Y>X) P(Y > X) , which is equivalent to P(Z>0) P(Z > 0) . Since ZN(0,2) Z \sim N(0, 2) , we standardize Z Z to a standard normal variable Z Z' :
Z=Z2N(0,1) Z' = \frac{Z}{\sqrt{2}} \sim N(0, 1)
Therefore,
P(Z>0)=P(Z2>0)=P(Z>0)=0.5 P(Z > 0) = P\left(\frac{Z}{\sqrt{2}} > 0\right) = P(Z' > 0) = 0.5

Conclusion:
12 \boxed{\frac{1}{2}}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.