Question

A telegraphic communication system transmits the signals dash and dot. Assume that $\frac{1}{3}$ of the dashes are changed to dots and $\frac{2}{5}$ of the dots are changed to dashes. Also suppose that the ratio of the transmitted dashes and dots is $3: 5$.
The probability that the transmitted signal is a dot if the signal received is a dot, is

• A. $\frac{6}{8}$
• B. $\frac{5}{8}$
• C. $\frac{4}{8}$
• D. $\frac{3}{8}$

Answer 1

Answer: (A)

A : event that a dash is received
B : event that dot is received
$H_1$: transmitted signal was a dash
$H _2$ : transmitted signal was a dot.
$P \left( B / H _1\right)=\frac{1}{3} ; P \left( A / H _2\right)=\frac{2}{5} $
$P \left( A / H _1\right)=\frac{2}{3} ; P \left( B / H _2\right)=\frac{3}{5}$
Signals transmitted dash $=3 k$ (say) and signals transmitted dot $=5 k$ (say)
$\Rightarrow P \left( H _1\right)=\frac{3}{8} ; P \left( H _2\right)=\frac{5}{8}$
$P ( A )= P \left( A \cap H _1\right)+ P \left( A \cap H _2\right)$
$= P \left( H _1\right) \cdot P \left( A / H _1\right)+ P \left( H _2\right) \cdot P \left( A / H _2\right)$

Now $P(A)=\frac{3}{8} \cdot \frac{2}{3}+\frac{5}{8} \cdot \frac{2}{5}=\frac{1}{2}$
$P ( B ) = P \left( B \cap H _1\right)+ P \left( B \cap H _2\right) $
$= P \left( H _1\right) \cdot P \left( B / H _1\right)+ P \left( H _2\right) \cdot P \left( B / H _2\right)$
$P ( B ) =\frac{3}{8} \cdot \frac{1}{3}+\frac{5}{8} \cdot \frac{3}{5}=\frac{1}{2}$
Now
$P \left( H _2 / B \right)=\frac{ P \left( H _2 \cap B \right)}{ P ( B )}=\frac{ P \left( H _2\right) \cdot P \left( B / H _2\right)}{ P ( B )}=\frac{\frac{5}{8} \cdot \frac{3}{5}}{\frac{1}{2}}=\frac{3}{8} \cdot \frac{2}{1}=\frac{3}{4}=\frac{6}{8}$