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 Dec 10, 2018
edited by shann0n  Dec 14, 2018
 #1
avatar+4442 
+1

\(\text{Let }A=\{\text{string starts with 10}\}\\ B=\{\text{string contains exactly 4 0s}\}\)

 

1)

\(P[0]=P[1]=\dfrac 1 2\\ \text{P[A] is the probability of selecting 10 out of the 4 possible 2 bit combos}\\ P[A] = \dfrac 1 4\\ P[B] = \dfrac{\binom{8}{4}}{2^8} = \dfrac{35}{128}\\ P[A \cap B] = P[\text{bits 1-2 are 10 and bits 3-8 have 3 0's}] = \\ \dfrac 1 4 \dfrac{\binom{6}{3}}{2^6}=\dfrac{5}{64}\)

 

\(P[A \cup B] = P[A]+P[B]-P[A \cap B]\\ P[A \cup B] = \dfrac 1 4 + \dfrac{35}{128}-\dfrac{5}{64} = \dfrac{57}{128}\)

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 Dec 10, 2018
 #2
avatar+4442 
+1

2)

\(P[1]=\dfrac 3 5=p,~P[0]=\dfrac 2 5\)

\(P[A]=\dfrac 3 5 \dfrac 2 5 = \dfrac{6}{25}\\ P[B] = \dbinom{8}{4}\left(\dfrac 3 5\right)^4\left(\dfrac 2 5\right)^4 = \dfrac{18144}{78125}\\ P[A \cap B] = \dfrac{6}{25} \dbinom{6}{3}\left(\dfrac 3 5\right)^3\left(\dfrac 2 5\right)^3 = \dfrac{5184}{78125}\)

\(P[A \cup B] = \dfrac{6}{25} + \dfrac{18144}{78125}-\dfrac{5184}{78125}=\dfrac{6342}{15625}\)

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 Dec 10, 2018
 #3
avatar+4442 
0

3)

\(P[A] = \left(1-\dfrac 1 2\right)\left(\dfrac 1 4\right) = \dfrac 1 8\\ P[B]= \text{.... are you supposed to be able to do this w/o software?}\\ \text{It's next to impossible by hand}\\ \text{unless I'm missing some major trick which is certainly possible}\)

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 Dec 10, 2018
edited by Rom  Dec 10, 2018

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