We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

#1**+1 **

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

1)

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

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

.Rom Dec 10, 2018

#2**+1 **

2)

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

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

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

.Rom Dec 10, 2018