Question

five (a) AUB-B 80f (1) AnB=A et B five 2.2.28. Let events A and B and sample space S be define the as the following intervals: S={x : 0 < x < 10} A={x : 0 < x <5) the Characterize the following events: as (a) AC (b) An B (c) AUB (d) AnB (e) ACUB (f) AC n B 2.2.29. A coin is tossed four timo

2.2.28
2.4.8
2.4.50

Answer 1

2.2.28

(a) A^c = { x ; x=0 or 5$\leq$x$\leq$10 }

(b) A$\cap$B = { x ; 3$\leq$x$<$5 }

(c) A$\cup$B = { x : 0$<$x$\leq$7 }

(d) B^c = { x : 0$\leq$x$<$3 or 7$<$x$\leq$10 }   A$\cap$B^c = { x : 0<x<3 }

(e) A^c$\cup$B = { x : x=0 or 3$\leq$x$\leq$10 }

(f) A^c$\cap$B^c = { x : x=0 or 7$<$x$\leq$10 }

2.4.8

P(A$\cup$B) = P(A) + P(B) - P(A$\cap$B)

P(A$\cup$B) $\leq$ 1 $\Rightarrow$ P(A) + P(B) - P(A$\cap$B) $\leq$ 1 $\Rightarrow$ P(A) + P(B) - 1 $\leq$ P(A$\cap$B) $\Rightarrow$ P(A$\cap$B) $\geq$ a+b-1

P(A|B) = P(A$\cap$B)/P(B)

P(A$\cap$B) $\geq$ a+b-1 $\Rightarrow$ P(A$\cap$B)/P(B) $\geq$ (a+b-1)/P(B) $\Rightarrow$ P(A|B) $\geq$ (a+b-1)/b

PROVED

2.4.50   

P(1 is sent) = 0.7 P(0 is sent) = 0.3

P(0 is received | 0 is sent) = 0.9 $\Rightarrow$ P(1 is received | 0 is sent) = 0.1

P(1 is received | 1 is sent) = 0.95 $\Rightarrow$ P(0 is received | 1 is sent) = 0.05

Required Probability = P(0 is sent | 1 is received)

According to Bayes theorem of conditional probability :-

$P(0 \;is \;sent\; | \;1 \;is \;received)=\frac{P(1\;is\;recieved\;|\;0\;is\;sent)P(0\;is\;sent)}{P(1\;is\;recieved\;|\;0\;is\;sent)P(0\;is\;sent)&plus;P(1\;is\;recieved\;|\;1\;is\;sent)P(1\;is\;sent)}$

  $=\frac{0.1\times0.3}{0.1\times0.3&plus;0.95\times0.7}=\frac{0.03}{0.695}\approx 0.043$