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# Coin Flipping Probability Mass Function & Expectation

$\begingroup$ The point is identical simpleton : The order of flip matters .
You are given four random variables $X_i, i = 1,2,3,4$, where each one stands for a flip of a coin, in ascending order of indices. You have to calculate the keep up quantity : $X = 8X_1 + 4X_2 + 2X_3 + X_4$, right ? This can have values between $0$ and $15$ .
For the foremost part : now, how is $X_i$ ? Well, it is $0$ or $1$ with probability half, so the expectation respect is just $E [ X ] = 8 \times 0.5 + 4 \times 0.5 + 2 \times 0.5+1 \times 0.5 = 7.5$, because the expectation of one toss is $0 \times 0.5 +1 \times 0.5 = 0.5$.

How about the mass officiate ? Well, each issue, from $0$ to $15$, has an equal probability of orgasm, because each one has a unique representation in binary, and the probability of heads and tails are the lapp. hence, it is merely $P ( one ) = \frac { 1 } { 16 }$, for $i=0,1, \ldots, 15$ ( there are sixteen possibilities in total, each occurs evenly credibly ) .
In the second question : The arithmetic mean of each $X_i$ nowadays is $0 \times \frac 13 + 1\times \frac 23 = \frac 23$, so then $E [ X ] = ( 8+4+2+1 ) \times\frac 23 = 10$ .
finally, here, every number has a unique binary star expansion, but this time the heads and tails do n’t have the same probability, so the mass affair changes, and it changes as follows : given the count of $1$ s in the expansion, so many tails are required for the issue to come, which happens with higher probability. so more the ones, more the probability .
so, we look at the binary star expansions :
1 ) The numbers having no $1$ sulfur is merely zero .
2 ) The numbers having precisely one $1$ are $2$, $4$, $8$, $1$ .
3 ) The numbers having precisely two $1$ s are $3$, $5$, $6$, $9$, $10$, $12$.

4 ) The numbers having precisely three ones are $7$, $11$, $13$, $14$ .
5 ) only $15$ has four ones .
therefore, we can write down the probabilities as follows :
1 ) For zero, the probability is $\frac { 1 } { 3^4 } = \frac { 1 } { 81 }$ .
2 ) For the numbers with only one $1$, the probability is $\frac { 1 } { 3^3 } \times \frac { 2 } { 3 } = \frac { 2 } { 81 }$ .
3 ) For the numbers having $2$ ones, the probability is $\frac { 1 } { 3^2 } \times \frac { 2^2 } { 3^2 } = \frac { 4 } { 81 }$.

4 ) For the numbers with three $1$ randomness, the probability is $\frac { 1 } { 3 } \times \frac { 2^3 } { 3^3 } = \frac { 8 } { 81 }$ .
5 ) For the numbers with all $1$ mho, the probability is $\frac { 2^4 } { 3^4 } = \frac { 16 } { 81 }$ .
hence, this is the probability distribution serve in the end :  P ( x ) = \begin { cases } \frac { 1 } { 81 } & x=0 \\ \frac { 2 } { 81 } & x=1,2,4,8 \\ \frac { 4 } { 81 } & x= 3,5,6,9,10,12 \\ \frac { 8 } { 81 } & x = 7,11,13,14 \\ \frac { 16 } { 81 } & x = 15 \\ \end { cases } 

reference : https://leowiki.com
Category : Economy