Envelope I contains 2 black and 3 red sheets. while envelope II contains 3 black and 4 red sheets. Out of these two envelopes, one is selected at random and the probability of choosing envelope I is double that of envelope II. If a red sheet is drawn from the selected envelope, then the probability that it has come from envelope II is:  

Key Points

 Given,
Envelope I contains 2 Black and 3 Red sheets while Envelope II contains 3 Black and 4 Red sheets. 
One envelope is selected at random and it is given probability of choosing envelope I is double that of envelope II. 

Let's denote the events of choosing envelopes I and II respectively as P(I) and P(II). 
As total probability is always 1, P(I) + P(II) = 1
Or, 2 P(II) + P(II) = 1 
Or, P(II) = 1/3 
Thus, P(I) = 2/3

The probability of choosing a red sheet from envelope I is P (R | I) = 3/5 and that from envelope II is P(R | II) = 4/7
According to Bayes' Theorem, the probability that the selected red sheet has come from envelope II is
= \( {{P(II)} \times {P(R | II)} \over P(I) \times P(R | I) + P(II) \times P(R|II)}\)

= \( {{1 \over 3} \times {4 \over 7} \over {{2\over 3} \times {3 \over 5} + {4 \over 7} \times {1 \over 3}}}\)

= \(10 \over 31\)

Hence, if a red sheet is drawn from the selected envelope, then the probability that it has come from envelope II is \(10 \over 31\).