__Problem :__

**Measuring information**

A.
Someone picks a name out
of a hat known to contain the names of 5 women and 3 men, and tells you a man
has been selected. How much information have they given you about the
selection?

Answer: -

There are 8 names to start with and knowing the
selection is a man narrows the choices down to 3 names. Using the formula from
lecture with N = 8 and M = 3, we've been given log

_{2}(8/3) bits of information.
Alternatively,
the probability of drawing a man's name is p

_{man}= 3/8, so the amount of information received is log_{2}(1/p_{man}) = log_{2}(1/(3/8)) = log_{2}(8/3).
B.
You're given a standard
deck of 52 playing cards that you start to turn face up, card by card. So far
as you know, they're in completely random order. How many new bits of
information do you get when the first card is flipped over? The fifth card? The
last card?

Answer: -

Before the first card was flipped over there are
52 choices for what we'll see on the first flip. Turning the first card over
narrows the choice down to a single card, so we've received log

_{2}(52/1) bits of information.
After
flipping over 4 cards, there are 48 choices for the next card, so flipping over
the fifth card gives us log

_{2}(48/1) bits of information.
Finally
if all but one card has been flipped over, we know ahead of time what the final
card has to be so we don't receive any information from the last flip. Using
the formula, there is only 1 "choice" for the card before the card is
flipped and we have the same "choice" afterwards, so, we receive log

_{2}(1/1) = 0 bits of information.
C.
X is an unknown N-bit
binary number (N > 3). You are told that the first three bits of X of 011.
How many bits of information about X have you been given?

Answer: -

Since we were told about 3 bits of X it would
make sense intuitively that we've been given 3 bits of information! Turning to
the formulas: there are 2

^{N}N-bit binary numbers and 2^{N-3}N-bit binary numbers that begin with 011. So we've been given log_{2}(2^{N}/2^{N-3}) = log_{2}(2^{3}) = 3 bits of information (whew!).
D. X is an unknown 8-bit binary number. You are
given another 8-bit binary number, Y, and told that the Hamming distance
between X and Y is one. How many bits of information about X have you been
given?

Answer: -

Before we learn about Y, there are 2

^{8}= 256 choices for X. If the Hamming distance between X and Y is one that means that X and Y differ in only one of their 8 bits, i.e., for a given Y there are only eight possible choices for X. So we've been given log_{2}(256/8) = 5 bits of information.
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