JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Probability XII

A signal which can be green or red with probability

\frac{4}{5}

and

\frac{1}{5}

respectively, is received by station A and then transmitted to station B. The probability of each station receiving the signal correctly is

\frac{3}{4}

. If the signal received at station

B

is green, then the probability that the original signal was green is

Answer: (C)

Solution

Let

G

= original signal is green,

E

= signal received at B is green.

Using Bayes' theorem:

P (G ∣ E) = \frac{P ( E ∣ G ) P ( G )}{P ( E )}

The signal reaches B as green in four ways:

Original green → A correct → B correct: $\frac{4}{5} \cdot \frac{3}{4} \cdot \frac{3}{4} = \frac{36}{80}$
Original green → A wrong → B wrong: $\frac{4}{5} \cdot \frac{1}{4} \cdot \frac{1}{4} = \frac{4}{80}$
Original red → A wrong → B correct: $\frac{1}{5} \cdot \frac{1}{4} \cdot \frac{3}{4} = \frac{3}{80}$
Original red → A correct → B wrong: $\frac{1}{5} \cdot \frac{3}{4} \cdot \frac{1}{4} = \frac{3}{80}$

P (E) = \frac{36 + 4 + 3 + 3}{80} = \frac{46}{80} = \frac{23}{40}

.

The probability that original was green and B receives green:

P (E \cap G) = \frac{36 + 4}{80} = \frac{40}{80} = \frac{1}{2}

Therefore:

P (G ∣ E) = \frac{40/80}{46/80} = \frac{40}{46} = \frac{20}{23}

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Single correct option