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Let x and y be independent random variables with binomial distribution B(10, 1 /3) and B(20, 1/3) respectively. E[x + y] is
  • a)
    5
  • b)
    10
  • c)
    15
  • d)
    30
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Let x and y be independent random variables with binomial distribution...
 E (x + y) = E (x) + E(y)

Expectation of binomial distribution is np if B(n, p) is the binomial distribution.
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Most Upvoted Answer
Let x and y be independent random variables with binomial distribution...
Understanding the Random Variables
Let x and y be independent random variables defined as follows:
- x ~ B(10, 1/3): This means x follows a binomial distribution with 10 trials and a probability of success of 1/3.
- y ~ B(20, 1/3): This means y follows a binomial distribution with 20 trials and a probability of success of 1/3.
Calculating Expected Values
To find E[x + y], we can utilize the property of expectation:
- E[x + y] = E[x] + E[y]
Now, we need to compute E[x] and E[y] individually.
Expected Value of x
For a binomial distribution B(n, p), the expected value is given by:
- E[x] = n * p
So for x:
- E[x] = 10 * (1/3) = 10/3 ≈ 3.33
Expected Value of y
Similarly, for y:
- E[y] = 20 * (1/3) = 20/3 ≈ 6.67
Combining the Expected Values
Now we can combine the expected values:
- E[x + y] = E[x] + E[y]
- E[x + y] = (10/3) + (20/3) = 30/3 = 10
Conclusion
Thus, the expected value of the sum of the two independent random variables x and y is:
- E[x + y] = 10
The correct answer is option b) 10.
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Community Answer
Let x and y be independent random variables with binomial distribution...
 E (x + y) = E (x) + E(y)

Expectation of binomial distribution is np if B(n, p) is the binomial distribution.
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Let x and y be independent random variables with binomial distribution B(10, 1 /3) and B(20, 1/3) respectively. E[x + y]isa)5b)10c)15d)30Correct answer is option 'B'. Can you explain this answer?
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