Probability of receiving a binary one

[Addez123]

Homework Statement

When transmitting ones and zeros through wires it encounters disturbance and can alter the recieved number from 0 to one and 1 to zero.

Given that you've recieved a one, whats the probability that one was actually sent?

Relevant Equations

The risk of sending 0 but receiving 1 is .01
The risk of sending 1 but receiving 0 is .02

The 0s and 1s are sent in the proportions:
Zeros .6
Ones .4

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

$$P(A) = \text{Chance of 1 being received} = .4 * .98 + .6 * .01 = .398$$
$$P(A \cap B) = \text{Chance 1 being sent and 1 being received} = .4 * .98 = .392$$
$$P(A|B) = P(A \cap B) / P(A) = .392 / .398 = .985$$

The correct answer is 147/148 ~= .9932

What am I doing wrong?

Also, am I supposed to use * for multiplication or is some other symbol prefered?

 

[BvU]

Hi,

The correct answer is 147/148 ~= .9932

Says who ? For the problem statement as you render it I don't see 147/148 popping up any way at all !

 

[Addez123]

The solution in the book says 147/148 :/

 

[PeroK]

The solution in the book says 147/148 :/

Your method looks right to me.

 

[PeroK]

The solution in the book says 147/148 :/

Try swapping the proportion of zeros and ones that are sent!

 

[haruspex]

Homework Statement:: When transmitting ones and zeros through wires it encounters disturbance and can alter the received number from 0 to one and 1 to zero.

Given that you've received a one, what's the probability that one was actually sent?
Relevant Equations:: The risk of sending 0 but receiving 1 is .01
The risk of sending 1 but receiving 0 is .02

The 0s and 1s are sent in the proportions:
Zeros .6
Ones .4

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

$$P(A) = \text{Chance of 1 being received} = .4 * .98 + .6 * .01 = .398$$
$$P(A \cap B) = \text{Chance 1 being sent and 1 being received} = .4 * .98 = .392$$
$$P(A|B) = P(A \cap B) / P(A) = .392 / .398 = .985$$

The correct answer is 147/148 ~= .9932

What am I doing wrong?

Also, am I supposed to use * for multiplication or is some other symbol prefered?

You got the right answer, but as a result of two errors that cancelled.
P(A|B) is the probability of event A given event B. The correct equation is $P(A|B) = \frac{P(A \cap B) }{ P(B)}$. Note the denominator.
The other error is that you defined A and B in such a way that your calculation should have yielded the probability of getting a 1 given that a 1 was sent, which is the converse of what is asked.