solve it please 3 (probability)
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THE FIRST TERM IN THE SEQUENCE IS p THT A RETURNS THE BALL SUCCESSFULLLY AND B MISSES IT SO (3/4)(1/5)...THE NEXT TERM IS A RETURNS SUCC. AND B RETURNS SUCC. AND A AGAIN RETURNS SUCC AND B MISSES IT....SO (3/4)(4/5)(3/4)(1/5)...AND SO ON UPTO INFINITY...... FOR SECOND PART ONLY 4 RETURNS R TO BE CONSIDERED SO 1ST 2 TERMS OF THE SEQUENCE COUNTS. GOT IT!!!!!!
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