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# ISI MStat PSB 2005 Problem 2 | Calculating probability using Binomial Distribution

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 !!

## Problem- ISI MStat PSB 2005 Problem 2

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.

### Prerequisites

Binomial Distribution

Binomial Expansion

Parity Check

## Solution :

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 .

## Food For Thought

Try to find P(X-Y=odd) under the same condition as given in the above problem .

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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 !!

## Problem- ISI MStat PSB 2005 Problem 2

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.

### Prerequisites

Binomial Distribution

Binomial Expansion

Parity Check

## Solution :

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 .

## Food For Thought

Try to find P(X-Y=odd) under the same condition as given in the above problem .

## Subscribe to Cheenta at Youtube

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