Question

9. A space probe near Neptune communicates with Earth using bit strings. Suppose that in its transmissions it sends a 1 one-third of the time and a 0 two-thirds of the time. When a 0 is sent, the probability that it is received correctly is 0.9, and the probability that it is received incorrectly (as a 1) is 0.1. When a 1 is sent, the probability that it is received correctly is 0.8, and the probability that it is received incorrectly (as a 0) is 0.2. (a) Find the probability that a 0 is received. (b) Use Bayes theorem to find the probability that a 0 was transmitted, given that a 0 was received.

Answer 1

Answer:

a) 0.6667 = 66.67% probability that a 0 is received.

b) 0.9 = 90% probability that a 0 was transmitted, given that a 0 was received.

Step-by-step explanation:

Bayes Theorem:

Two events, A and B.

$P(B|A) = \frac{P(B)*P(A|B)}{P(A)}$

In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.

(a) Find the probability that a 0 is received.

0.9 of 2/3(0 received when a 0 is sent).

0.2 of 1/3(0 received when a 1 is sent). So

$p = \frac{0.9*2}{3} + \frac{0.2*1}{3} = 0.6667$

0.6667 = 66.67% probability that a 0 is received.

(b) Use Bayes theorem to find the probability that a 0 was transmitted, given that a 0 was received.

Event A: 0 received

Event B: 0 transmitted.

0.6667 = 66.67% probability that a 0 is received, which means that $P(A) = 0.6667$

A zero is transmitted two-thirds of time, which means that $P(B) = 0.6667$

When a 0 is sent, the probability that it is received correctly is 0.9, which means that $P(B|A) = 0.9$

So

$P(B|A) = \frac{P(B)*P(A|B)}{P(A)} = \frac{0.6667*0.9}{0.6667} = 0.9$

0.9 = 90% probability that a 0 was transmitted, given that a 0 was received.