A 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 is 3 / 4. If the signal received at station B is given, then the probability that the original signal is green, is

Question

A 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 is 3 / 4. If the signal received at station B is given, then the probability that the original signal is green, is (A) (3/5) (B) (6/7) (C) (20/23) (D) (9/20)

Tardigrade Admin · Accepted Answer

From the tree diagram, it follows that <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/e3251ff882a8fed37b52ecaf783d6b4e-.png" /> ∴ P(BG ∩ G)=(5/8) × (4/5)=(1/2)[∵ P(BG ∩ G) =P((BG/G)) × P(G)] text Now, P(G / BG)=(P(BG ∩ G)/P(BG))=(1/2) × (80/46)=(20/23)

$\frac{3}{5}$

$\frac{6}{7}$

$\frac{20}{23}$

$\frac{9}{20}$

From the tree diagram, it follows that

$∴ P (B_{G} \cap G) = \frac{5}{8} \times \frac{4}{5} = \frac{1}{2} [∵ P (B_{G} \cap G) = P (\frac{B _{G}}{G}) \times P (G)]$
$Now, P (G / B_{G}) = \frac{P ( B _{G} \cap G )}{P ( B _{G} )} = \frac{1}{2} \times \frac{80}{46} = \frac{20}{23}$

53​

76​

2320​

209​

From the tree diagram, it follows that ∴P(BG​∩G)=85​×54​=21​[∵P(BG​∩G)=P(GBG​​)×P(G)] Now, P(G/BG​)=P(BG​)P(BG​∩G)​=21​×4680​=2320​

$\frac{3}{5}$

$\frac{6}{7}$

$\frac{20}{23}$

$\frac{9}{20}$

From the tree diagram, it follows that

$∴ P (B_{G} \cap G) = \frac{5}{8} \times \frac{4}{5} = \frac{1}{2} [∵ P (B_{G} \cap G) = P (\frac{B _{G}}{G}) \times P (G)]$
$Now, P (G / B_{G}) = \frac{P ( B _{G} \cap G )}{P ( B _{G} )} = \frac{1}{2} \times \frac{80}{46} = \frac{20}{23}$