Box A contains 25 slips of which 19 marked Rs 1 and other are marked Rs 5 each. Box B contains 50 slips of which 45 marked Rs 1 each and others are marked Rs 13 each. Slips of both boxes are poured into a third box and reshuffled. A slip is drawn at random. What is the probability that it is marked other than Rs 1?

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Solution

Total number of slips in a box, n (S) = 25 + 50 = 75

Since, the total number of slips that are marked as Rs.1 is 19+45=64.
Thus, it is clear that there are 75-64=11 slips which are marked other than Rs 1.
$∴$ Required probability $= \frac{N u m b e r o f s l i p s o t h e r t h a n R s 1}{T o t a l n u m b e r o f s l i p s} = \frac{11}{75}$

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Total number of slips in a box, n (S) = 25 + 50 = 75 Since, the total number of slips that are marked as Rs.1 is 19+45=64.Thus, it is clear that there are 75-64=11 slips which are marked other than Rs 1. ∴ Required probability =Number of slips other than Rs 1Total number of slips=1175