Answer:
- 45/1024
- 1/4
- 15/128
- 193/512
- 9/512
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
