# Probability of original signal

Signal which can be green or red with probability $4/5$ and $1/5$ respectively, is received by station A and then transmitted to station B. The probability of each station receiving the signal correctly $3/4$. If the Signal received at station B is green, then the probability that the original signal was green is

In this I think we have to use Bayes' theorem . but could not able to apply it .

## 3 Answers

The possible sequences ending in G are RRG, RGG, GGG, GRG. Here RRG means signal emitted red, received by A as red, received by B as green.

The probabilities are
$p(RRG)=\frac15 \frac34 \frac14 = \frac{3}{80}$
$p(RGG)=\frac15 \frac14 \frac34 = \frac{3}{80}$
$p(GGG)=\frac45 \frac34 \frac34 = \frac{36}{80}$
$p(GRG)=\frac45 \frac14 \frac14 = \frac{4}{80}$

The probability that G was emitted if G is received at B (sequences GGG and GRG) is
$\frac{36+4}{36+4+3+3}=\frac{40}{46}=\frac{20}{23}.$

$\mathbb{P}(\text{the signal transmitted was green | B received a green signal}) = \frac{\mathbb{P}(\text{the signal transmitted was green }\cap\text{ B received a green signal})}{\mathbb{P}(\text{B received a green singal })}$

Here, $\mathbb{P}(\text{the signal transmitted was green }\cap\text{ B received a green signal}) = \mathbb{P}(\text{the signal transmitted was green }\cap\text{ A received a green signal}\cap\text{ B received a green signal})+\mathbb{P}(\text{the signal transmitted was green }\cap\text{ A received a red signal}\cap\text{ B received a green signal}) = \frac{4}{5}\cdot \frac{3}{4}\cdot \frac{3}{4} + \frac{4}{5}\cdot \frac{1}{4}\cdot \frac{1}{4} = \frac{1}{2}$

Next, ${\mathbb{P}(\text{B received a green singal })} = \mathbb{P}(\text{the signal transmitted was green }\cap\text{ B received a green signal}) + \mathbb{P}(\text{the signal transmitted was red }\cap\text{ B received a green signal})$

The former term here has already been evaluated to be $\frac{1}{2}$. The latter term can be similarly evaluated as follows: $\mathbb{P}(\text{the signal transmitted was red }\cap\text{ B received a green signal}) = \mathbb{P}(\text{the signal transmitted was red }\cap\text{ A received a red signal}\cap\text{ B received a green signal}) + \mathbb{P}(\text{the signal transmitted was red }\cap\text{ A received a green signal}\cap\text{ B received a green signal}) = \frac{1}{5}\cdot\frac{3}{4}\cdot \frac{1}{4} + \frac{1}{5}\cdot\frac{1}{4}\cdot \frac{3}{4} = \frac{3}{40}$

That is, ${\mathbb{P}(\text{B received a green singal })} = \frac{23}{40}$

Thus, in accordance with the first equation, $\mathbb{P}(\text{the signal transmitted was green | B received a green signal}) = \dfrac{\frac{1}{2}}{\frac{23}{40}} = \frac{20}{23}$

• How you have written $\mathbb{P}(\text{the signal transmitted was green }\cap\text{ B received a green signal}) = \mathbb{P}(\text{the signal transmitted was green }\cap\text{ A received a green signal}\cap\text{ B received a green signal})+\mathbb{P}(\text{the signal transmitted was green }\cap\text{ A received a red signal}\cap\text{ B received a green signal})$ Apr 25, 2017 at 17:35
• @user123733 it's because in the right hand side of this equation, the two events, the corresponding probabilities of which are being added, are mutually exclusive and their union gives the event probability of which is on the left hand side. Apr 25, 2017 at 17:44

Direct Application of Baye's Theorem :

P(Og)= Probability that original signal is Green.

P(G)=Probability that received signal is green.

(Red gets represented by P(Or) and P(R) respectively )

P(Og|G)=(P(Og) P(G|Og))/(P(Og) P(G|Og) + P(Or) P(G|Or))

P(Og)=4/5 P(Or)=1/5 P(G|Og)=(3/4*3/4+1/4*1/4) P(G|Or)= (3/4*1/4+1/4*3/4)

P(Og|G) [after putting above values]= 20/23

(Note : P(G|Og) means Green received by B given that original was Green. Which can happen when both Stations Rightly receive the signal or both the station receive the signal wrong. So : (3/4*3/4+1/4*1/4)

P(G|Or) means Green received by B given that original signal was Red which can happen when either of the stations receive it faulty but other station receive correct. So : (3/4*1/4+1/4*3/4) )