binomial distribution: verify formula b(x + 1 ;,n,theta)=(theta* (n-x)/(x+1)*(1-theta

Ambika

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when calculating all the values of a Binomial Distribution the work and usually be simplified by first calculating b(0,n,theta)
and then using the recursion formula
b(x + 1 ;,n,theta)=(theta* (n-x)/(x+1)*(1-theta) ) *b(x;n,theta)verify this formula and use it too calculate the values of the Binomial Distribution with n equal to 7 and theta is equal to 0.25
 
when calculating all the values of a Binomial Distribution the work and usually be simplified by first calculating b(0,n,theta)
and then using the recursion formula
b(x + 1 ;,n,theta)=(theta* (n-x)/(x+1)*(1-theta) ) *b(x;n,theta)verify this formula and use it too calculate the values of the Binomial Distribution with n equal to 7 and theta is equal to 0.25
What are your thoughts?

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When calculating all the values of a binomial distribution, the work can usually be simplified by first calculating

. . . . .\(\displaystyle b(0;\, n,\, \theta)\)

...and then using the recursion formula:

. . . . .\(\displaystyle b(x\, +\, 1;\, n,\, \theta)\, =\, \left(\, \dfrac{\theta\, (n\, -\, x)}{(x\, +\, 1)(n\, -\, \theta)}\, \right)\, \cdot\, b(x;\, n,\, \theta)\)

Verify this formula, and use it to calculate the values of the binomial distribution with n = 7 and \(\displaystyle \, \theta\, =\, 0.25.\)
Note: In the above, I am assuming that \(\displaystyle \, \theta\, \) stands for the probability of a success in a single trial, n stands for the number of trials, and x stands for the exact number of desired successes.

How far have you gotten in doing the verification? Where are you stuck? Please be complete: you may be nearly done already! Thank you! ;)
 
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