Question

In: Statistics and Probability

Show (prove, demonstrate, whatever) that the number of ways to obtain r "successes" (1's) in a...

Show (prove, demonstrate, whatever) that the number of ways to obtain r "successes" (1's) in a series of n binary outcomes is equal to the combination nCr. For example, the number of ways to obtain 2 successes (1's) in a sequence of 3 binary digits is 3; or in a sequence of 4 bits is 6. This came up in our derivation of the binomial probability distribution; it is also covered in chapter 5.

Expert Solution

Solution: Consider a set of 2 independent Bernoullian trials in which probability 'p' of success in any trial is constant for each trial, then q=1-p, is the probability of failure in any trial.

The probability of 1 success and consequently (2-1) failures in 2 independent trials, in a specified order (say) SF (where S represents success and F represents failure) is given by the compound probability theorem by the expression:

But 1 success in 2 trials can occur in two ways that is like SF or FS and the probaility for each of these ways is same, viz., . Hence the probability of 1 success in 2 trials in any order is given by the addition theorem of probability by the expression

Now if we consider a set of 3 independent Bernoullian trials in which probability 'p' of success in any trial is constant for each trial, then q=1-p, is the probability of failure in any trial.

The probability of 1 success and consequently (3-1) failures in 3 independent trials, in a specified order (say) SFF (where S represents success and F represents failure) is given by the compound probability theorem by the expression:

But 1 success in 3 trials can occur in two ways that is like SFF, FSF or FFS and the probaility for each of these ways is same, viz., . Hence the probability of 1 success in 3 trials in any order is given by the addition theorem of probability by the expression

Therefore using the same logic, if we onsider a set of n independent Bernoullian trials in which probability 'p' of success in any trial is constant for each trial, then q=1-p, is the probability of failure in any trial.

The probability of x success and consequently (n-x) failures in n independent trials will be:

Hope it helps

orchestra answered 1 year ago

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Question

Show (prove, demonstrate, whatever) that the number of ways to obtain r "successes" (1's) in a...

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Expert Solution

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