There are N coins in a box in which M are balanced coin and remaining are unbalanced. When balanced coin is tossed has probability `(1)/(2)` and for unbalanced coin probability is `(2)/(3)` Now one coin is selected from box at random and it is tossed twice, It is known that head on first toss and tail on second toss is obtained. Then prove that probability of an event that selected coin is balanced is `(9M)/(8N + M)`.
KUMAR PRAKASHAN|Exercise Practice Paper - 3 (Section - D)|1 Videos
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KUMAR PRAKASHAN-PROBABILITY-Practice Paper - 13 (Section - D (Answer the following questions))
