Suppose that you pick a bit string from the set of all bit strings of length ten. Find the probability that a) the bit string has exactly two 1s; b) the bit string begins and ends with 0; c) the bit string has the sum of its digits equal to seven; d) the bit string has more 0s than 1s; e) the bit string has exactly two 1s, given that the string begins with a 1.

Step-by-step explanation:

There are 2^10 = 1024 bit strings of length 10.

a) There are 10C2 = 45 ways to have exactly two 1-bits in 10 bits

  p(2 1-bits) = 45/1024

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b) Of the four (4) possibilities for beginning and ending bits (00, 01, 11, 10), exactly one (1) of those is 00.

  p(b0=0 & b9=0) = 1/4

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c) There are 10C7 = 120 ways to have seven 1-bits in the bit string.

  p(7 1-bits) = 120/1024 = 15/128

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d) ∑10Ck {for k=0 to 4} = 386 is the total of the number of ways to have 0, 1, 2, 3, or 4 1-bits in the string. If there are more than that, there won't be more 0-bits than 1-bits

  p(more 0 bits) = 386/1024 = 193/512

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e) The string will have two 1-bits if it starts with 1 and there is a single 1-bit among the other 9 bits. There are 9 ways that can happen, among the 512 ways to have 9 remaining bits.

  p(2 1-bits | first is a 1-bit) = 9/512

2(2x + 3) = -4 + 2(2x + 5)

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 6

4x - 6 + 6 = 4x - 6 + 6

4x = 4x

x = x

unlimited solutions.