Expected Value of X in Binomial Trials

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In a series of three binomial trials with a success probability of p=0.4, the expected value of the random variable X, representing outcomes as base 2 numbers, is calculated to be E[X] = 2.80. When the probability of success is adjusted to P[S]=0.5, the expected value changes to E[X] = 3.5. The discussion clarifies that the trials should be referred to as Bernoulli trials, which have only two outcomes. The expected value is derived using the definition of expectation, incorporating the probabilities of each outcome. This approach effectively demonstrates how to calculate expected values in this context.
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In a series of three binomial trials with p=.4, a random variable, X, is assigned to the outcomes to form a base 2 number, with 1 associated with success(S), and 0 associated with failure(F). For example, SFS->101=5.

A) Find the expected value of X, E[X].
(Answer 2.80)

B) Find the expected value of X if P=.5
(Answer 3.5)
 
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I think you mean 3 bernoulli trials and not binomial trials. Remember bernoulli trials have just 2 outcomes (1 and 0 in your question).
Since you know the 3 digit number (from the SFS or other combinations) just use definition of Expectation.
E[X] = Summation (x*P(x))
In your case the x would be 1*2^n and 0*2^n(which is 0 so can ignore)...n is the power corresponding to the placing of the number.
As for a number at the unit's place, you would have 1*2^0.
For Bernoulli the expectation is just p and then use the definition to obtain the result.

Hope this helps.
 
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