Coin Flipping Probability Mass Function & Expectation

$ \begingroup $ The target is very simple : The regulate of flip matters .
You are given four random variables $ X_i, i = 1,2,3,4 $, where each one stands for a chuck of a mint, in ascending order of indices. You have to calculate the succeed measure : $ X = 8X_1 + 4X_2 + 2X_3 + X_4 $, properly ? This can have values between $ 0 $ and $ 15 $ .
For the first contribution : now, how is $ X_i $ ? Well, it is $ 0 $ or $ 1 $ with probability half, so the expectation value is good $ E [ X ] = 8 \times 0.5 + 4 \times 0.5 + 2 \times 0.5+1 \times 0.5 = 7.5 $, because the arithmetic mean of one chuck is $ 0 \times 0.5 +1 \times 0.5 = 0.5 $.

How about the mass function ? Well, each number, from $ 0 $ to $ 15 $, has an equal probability of come, because each one has a unique representation in binary, and the probability of heads and tails are the lapp. hence, it is barely $ P ( iodine ) = \frac { 1 } { 16 } $, for $ i=0,1, \ldots, 15 $ ( there are sixteen possibilities in total, each occurs evenly probably ) .
In the second question : The expectation of each $ X_i $ immediately is $ 0 \times \frac 13 + 1\times \frac 23 = \frac 23 $, sol then $ E [ X ] = ( 8+4+2+1 ) \times\frac 23 = 10 $ .
finally, here, every number has a alone binary expansion, but this time the heads and tails do n’t have the lapp probability, so the batch affair changes, and it changes as follows : given the number of $ 1 $ sulfur in the expansion, so many tails are required for the act to come, which happens with higher probability. then more the ones, more the probability .
then, we look at the binary expansions :
1 ) The numbers having no $ 1 $ sulfur is alone nothing .
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 ) alone $ 15 $ has four ones .
hence, 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 $ s, the probability is $ \frac { 1 } { 3 } \times \frac { 2^3 } { 3^3 } = \frac { 8 } { 81 } $ .
5 ) For the numbers with all $ 1 $ randomness, the probability is $ \frac { 2^4 } { 3^4 } = \frac { 16 } { 81 } $ .
hence, this is the probability distribution function 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 } $ $

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