Question

2. For this question, use the joint probability table below. It involves three random variables: A € {21,22,23}, B € {b,–b}, and C € {c-c} b b b b ai 0.012 0.09 0.049 0.063 A2 0.024 0.072 0.098 0.084 a3 0.084 0.018 0.343 0.063 a. Reduce this to the joint probability tables between B and C only b. Calculate P(B= b |C=n) and P(C=C).

Answer 1

Answer:

Given that,

$\rightarrow$ Use the joint probability table below. It involves three variables:

Ae {(1, 02, a3, Be{6, -6}, and Ce{c, -c}

	c		-c
	b	-b	b	-b
a1	0.012	0.09	0.049	0.063
a2	0.024	0.072	0.098	0.084
a3	0.084	0.018	0.343	0.063

Probabilities are calculated as follows:

b & c=0.012+0.024+0.084=0.12

b & c=0.049+0.098+0.343=0.49

-b & c=0.09+0.072+0.018=0.18

-b & -c=0.063+0.084+0.063=0.21

(a).

The joint probability table of (B,C) is as follows:

B/C	c	-c	Total
b	0.12	0.49	0.61
-b	0.18	0.21	0.39
Total	0.30	0.70	1

The marginals of B and C are as follows:

B	b	-b
P(B)	0.61	0.39

C	c	-c
P(C)	0.30	0.70

(b).

$\rightarrow$

$P(B=-b|C=-c)=\frac{P(B=-b\cap C=-c)}{P(C=-c)}$

[Since,   ]

PAB) P(AB) - PAN B) P(B)

$=\frac{0.21}{0.70}$

$=\frac{21}{70}$

$=0.3$

$\rightarrow$ P(C=c)=0.30 [Since, from marginal table of C]