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 .
 A: $\mathbb{P}(\text{the signal transmitted was green | B received a green signal}) = \frac{\mathbb{P}(\text{the signal transmitted was green }\cap\text{ B received a green signal})}{\mathbb{P}(\text{B received a green singal })}$
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, $\mathbb{P}(\text{the signal transmitted was green | B received a green signal}) = \dfrac{\frac{1}{2}}{\frac{23}{40}} = \frac{20}{23}$
A: 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
$\frac{36+4}{36+4+3+3}=\frac{40}{46}=\frac{20}{23}.$
A: 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)
)
