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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 .

Content

[hide]

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 .

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