The probability that symbol j is sent and symbol k is received

[SamTaylor]

Homework Statement

A Communication system transmits signals labeled 1, 2, and 3. The probability
that symbol j is sent and symbol k is received is listed in the table for each
pair (j,k) of sent and received symbols. For example, the probability is 0.12
that a 1 is sent and, owning to noise in the channel 3 is received.

Calculate the probability that the symbol k was sent, given that symbol k is
received, for k = 1,2,3, and calculate the probability of error incurred in
using this system. An Error is defined as the reception of any symbol other
than the one transmitted.

Homework Equations

1
$$P(B|A)=\frac{P(A|B)*P(B)}{P(A)}$$
2
$$P(M)=P(A)*P(M|A) + P(B)*P(M|B) + P(C)*P(M|C)$$

The Attempt at a Solution

Problem 1: Symbol k was sent, given that k is received
$$P = P_{11} + P_{22} + P_{33}$$

Problem 2: probability of Error
$$P_e = P - 1$$

Or do the numbers inside the table represent conditional probabilitys?
P(received|send) ... for example P(3|1) = 0.12

 

[EnumaElish]

I think you need to calculate p[1|1], p[2|2], p[3|3] separately using eqs. 1 and 2.

 

[SamTaylor]

I thought about that but how do i get P(K) or P(J) ?

 

[HallsofIvy]

P(k)= P(k|1)+ P(k|2)+ P(k|3)

 

[SamTaylor]

$$P(K_m) = P(J_n)*P(K_m|J_1) + P(J_n)*P(K_m|J_2) + P(J_n)*P(K_m|J_3)$$

$$P(K_1) = 0.10 + 0.07 + 0.10 = 0.27$$
$$P(K_2) = 0.06 + 0.15 + 0.15 = 0.36$$
$$P(K_3) = 0.12 + 0.05 + 0.20 = 0.37$$

a)
$$P(J_1|K_1) = \frac{P(J_1)*P(K_1|J_1)}{P(K_1)} = \frac{0.10}{0.27} = 0.37$$
$$P(J_2|K_2) = \frac{P(J_2)*P(K_2|J_2)}{P(K_2)} = \frac{0.15}{0.36} = 0.4167$$
$$P(J_3|K_3) = \frac{P(J_3)*P(K_3|J_3)}{P(K_3)} = \frac{0.20}{0.37} = 0.541$$

$$P(J_n) = P(K_1)*P(K_1|J_n) + P(K_2)*P(K_2|J_n) + P(K_3)*P(K_3|J_n)$$
$$P(J_1) = 0.10 + 0.06 + 0.12 = 0.28$$
$$P(J_2) = 0.07 + 0.15 + 0.05 = 0.27$$
$$P(J_3) = 0.10 + 0.15 + 0.20 = 0.45$$

$$P(K_1)*P(J_1|K_1) = 0.3700 * 0.28 = 0.1036$$
$$P(K_2)*P(J_2|K_2) = 0.4167 * 0.27 = 0.1125$$
$$P(K_3)*P(J_3|K_3) = 0.5410 * 0.20 = 0.1082$$

b)
$$Q = P(K_1)*P(J_1|K_1) + P(K_2)*P(J_2|K_2) + P(K_3)*P(J_3|K_3) = 0.3243$$
$$P_e = 1-Q = 0.6757$$

Is that correct? If yes how do i know that the values inside the
table are the values for P(J)*P(K|J) and not only for P(K|J)?

 

[SamTaylor]

Ok, the last post was nonsense... sorry