This is a very beautiful sample problem from ISI MStat PSB 2005 Problem 2 based on finding probability using the binomial distribution. Let's give it a try !!
Let \(X\) and \(Y\) be independent random variables with X having a binomial distribution with parameters 5 and \(1 / 2\) and \(Y\) having a binomial distribution with parameters 7 and \(1 / 2 .\) Find the probability that \(|X-Y|\) is even.
Binomial Distribution
Binomial Expansion
Parity Check
Given \( X \sim \) Bin(5,1/2) and \( Y \sim \) Bin(7,1/2) , and they are independent .
Now , we have to find , \( P(|X-Y|=even ) \)
\( |X-Y| \)= even if both X and Y are even or both X and Y are odd .
Therefore \( P(|X-Y|=even )=P(X=even,Y=even) + P(X=odd , Y=odd) \)
P(X=even , Y= even ) =\( ( {5 \choose 0} {(\frac{1}{2})}^5 + {5 \choose 2} {(\frac{1}{2})}^5 + \cdots + {5 \choose 4} {(\frac{1}{2})}^5 )( {7 \choose 0} {(\frac{1}{2})}^7 + {7 \choose 2} {(\frac{1}{2})}^7 + \cdots + {7 \choose 6} {(\frac{1}{2})}^7)\)
=\( ({(\frac{1}{2})}^5 \times \frac{2^5}{2})({(\frac{1}{2})}^7 \times \frac{2^7}{2}) \)
= \(\frac{1}{4} \)
Similarly , one can find P(X=odd , Y=odd ) which is coming out to be \( \frac{1}{4} \) .
Hence , P(|X-Y|) = 14+1/4 = 1/2 .
Try to find P(X-Y=odd) under the same condition as given in the above problem .
This is a very beautiful sample problem from ISI MStat PSB 2005 Problem 2 based on finding probability using the binomial distribution. Let's give it a try !!
Let \(X\) and \(Y\) be independent random variables with X having a binomial distribution with parameters 5 and \(1 / 2\) and \(Y\) having a binomial distribution with parameters 7 and \(1 / 2 .\) Find the probability that \(|X-Y|\) is even.
Binomial Distribution
Binomial Expansion
Parity Check
Given \( X \sim \) Bin(5,1/2) and \( Y \sim \) Bin(7,1/2) , and they are independent .
Now , we have to find , \( P(|X-Y|=even ) \)
\( |X-Y| \)= even if both X and Y are even or both X and Y are odd .
Therefore \( P(|X-Y|=even )=P(X=even,Y=even) + P(X=odd , Y=odd) \)
P(X=even , Y= even ) =\( ( {5 \choose 0} {(\frac{1}{2})}^5 + {5 \choose 2} {(\frac{1}{2})}^5 + \cdots + {5 \choose 4} {(\frac{1}{2})}^5 )( {7 \choose 0} {(\frac{1}{2})}^7 + {7 \choose 2} {(\frac{1}{2})}^7 + \cdots + {7 \choose 6} {(\frac{1}{2})}^7)\)
=\( ({(\frac{1}{2})}^5 \times \frac{2^5}{2})({(\frac{1}{2})}^7 \times \frac{2^7}{2}) \)
= \(\frac{1}{4} \)
Similarly , one can find P(X=odd , Y=odd ) which is coming out to be \( \frac{1}{4} \) .
Hence , P(|X-Y|) = 14+1/4 = 1/2 .
Try to find P(X-Y=odd) under the same condition as given in the above problem .
