Question

Find p[b] in each case: (a) events a and b are a partition and p[a] = 3p[b]. (b) for events a and b, p[a ∪ b] = p[a] and p[a ∩ b] = 0. (c) for events a and b, p[a ∪ b] = p[a]− p[b].

Answer 1

Answer:

(a) $p[b]=\frac{1}{3}\times p[a]$.

(b) p[b]=0

(c) $p[b]=\frac{1}{2}\times p[a\cap b]$

Step-by-step explanation:

(a)

$p[a]=3p[b]$

Divide both sides by 3.

$\frac{1}{3}\times p[a]=p[b]$

Therefore $p[b]=\frac{1}{3}\times p[a]$.

(b)

$p[a\cup b]=p[a]$

$p[a]+p[b]-p[a\cap b]=p[a]$         $[\because P(A\cup B)=P(A)+P(B)-P(A\cap B)]$

$p[a]+p[b]-0=p[a]$                         $[\because p[a\cap b]=0]$

$p[b]=p[a]-p[a]$

$p[b]=0$

Therefore $p[b]=0$.

(c)

$p[a\cup b]=p[a]-p[b]$

$p[a]+p[b]-p[a\cap b]=p[a]-p[b]$   $[\because P(A\cup B)=P(A)+P(B)-P(A\cap B)]$

$p[b]-p[a\cap b]+p[b]=0$

$2p[b]=p[a\cap b]$

$p[b]=\frac{1}{2}\times p[a\cap b]$

Therefore $p[b]=\frac{1}{2}\times p[a\cap b]$.

Answer 2

Answer:

Case(a): $p[b]=\frac{1}{3}p[a]$

Case(b): $p[b]=0$

Case(c):  $p[b]=\frac{1}{2}p[a\cap b]$

Step-by-step explanation:

Given

(a) events a and b are a partition and p[a] = 3p[b].

(b) for events a and b, p[a ∪ b] = p[a] and p[a ∩ b] = 0.

(c) for events a and b, p[a ∪ b] = p[a]− p[b].

we have to find the p[b] in each case:

Case (a): events a and b are a partition and p[a] = 3p[b].

gives $p[b]=\frac{1}{3}p[a]$

Case (b):  for events a and b, p[a ∪ b] = p[a] and p[a ∩ b] = 0.

$p[a\cup b]=p[a]$ ⇒$p[a]+p[b]-p[a\cap b]=p(a)$ ⇒ $p[b]=0$  ∵  p[a ∩ b] = 0.

Case(3):  for events a and b, p[a ∪ b] = p[a]− p[b].

p[a ∪ b] = p[a]− p[b]

⇒ $p[a]+p[b]-p[a\cap b]=p[a]-p[b]$

⇒ $2p[b]=p[a\cap b]$

⇒  $p[b]=\frac{1}{2}p[a\cap b]$