ProbabilityHard
Question
A bag contains 7 tickets marked with the numbers 0, 1, 2, 3, 4, 5, 6 respectively. A ticket is drawn & replaced. Then the chance that after 4 drawings the sum of the numbers drawn is 8 is :
Options
A.165/2401
B.149/2401
C.3/49
D.none
Solution
Total cases x8 : (x0 + x1 + ...x6)4 = 
= (1 - x7)4 (1 - x)-4
= (1 - 2x7)2 (1 - x)-4 = (1 - 4x7) (1 - x)-4
Total ways = 7 4 4 + 8 - 1C8 - 4 4 + 1 - 1C1 = 11C8 - 4 × 4
= 165 - 16 = 149.
P =
= (1 - x7)4 (1 - x)-4
= (1 - 2x7)2 (1 - x)-4 = (1 - 4x7) (1 - x)-4
Total ways = 7 4 4 + 8 - 1C8 - 4 4 + 1 - 1C1 = 11C8 - 4 × 4
= 165 - 16 = 149.
P =
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