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Only 1 in 10 persons opt for a middle berth in "sleeper class" of Indian railways. What is the probability of getting a middle berth in sleeper compartment?

In every coupe, there are 2 Middle Berths out of 8 total berths.

Let me try to solve: Simple probability of getting a middle berth is 2/8 = 0.25, but actual probability, I suspect, is very high (as only 1 in 10 opt for it, 9/10 do not opt for it).

$$ P(M \mid D) = \frac{P(D \mid M)\, P(M)}{P(D)} = \frac{0.25 \times 0.9}{P(D)} $$

I am stuck up here, what is $P(D)$?

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  • $\begingroup$ Will this question be ever approved, @Tim? $\endgroup$
    – Felix Bast
    Aug 29, 2019 at 12:26

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By the law of total probability

$$ P(D) = P(D|M)\,P(M) + P(D|\neg M)\,P(\neg M) $$

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  • $\begingroup$ (0.25 x 0.9) / {(0.25 x 0.9) + (0.75 x 0.1)} = 0.75, 75% chance. Is it correct? Thanks lot Tim $\endgroup$
    – Felix Bast
    Aug 30, 2019 at 7:29

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