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← IIT-JEE 2010 · Paper 2
MathematicsSingle correct+5 / -2Question 25 of 57

IIT-JEE 2010 · Paper 2, Mathematics question 25

A signal which can be green or red with probability 45\dfrac45 and 15\dfrac15 respectively, is received by station A and then transmitted to station B. The probability of each station receiving the signal correctly is 34\dfrac34. If the signal received at station B is green, then the probability that the original signal was green is

Choose one option
Answer and worked solution

Official answer: C

B receives green if the signal arrives correctly twice or is flipped twice.

  • Original green (45\frac45): (34)2+(14)2=1016\left(\frac34\right)^2+\left(\frac14\right)^2=\frac{10}{16}.
  • Original red (15\frac15): exactly one flip, 2⋅34⋅14=6162\cdot\frac34\cdot\frac14=\frac6{16}.

Bayes.
P(green∣B green)=45⋅101645⋅1016+15⋅616=4046=2023.P(\text{green}\mid B\text{ green})=\frac{\frac45\cdot\frac{10}{16}}{\frac45\cdot\frac{10}{16}+\frac15\cdot\frac6{16}}=\frac{40}{46}=\frac{20}{23}.

(C)

Topics:Bayes' theorem
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