Four tickets marked 00, 01,10, 11 respectively are placed in a bag. A ticket is drawn at random five times, being replaced each time . the probability that the sum of the numbers on tickets thus drawn is 23?
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17
4 tickets can be drawn 5x in 4⁵ ways
sum= 23
= co-eff of x²³ in (x⁰ + x¹ + x¹⁰ + x¹¹)⁵
= co-eff of x²³ in (1 + x + x¹⁰ + x¹¹)⁵
= co-eff of x²³ in (1 + x)⁵ (1 + x¹⁰)⁵
= co-eff of x²³ in (1 + 5x + 10x² + 10x³ + 5x⁴+ x⁵) (1 + 5x¹⁰ + 10x²⁰+ ...)
= 10 x 10
= 100
probability = 100/ 4⁵
= 25 /256
sum= 23
= co-eff of x²³ in (x⁰ + x¹ + x¹⁰ + x¹¹)⁵
= co-eff of x²³ in (1 + x + x¹⁰ + x¹¹)⁵
= co-eff of x²³ in (1 + x)⁵ (1 + x¹⁰)⁵
= co-eff of x²³ in (1 + 5x + 10x² + 10x³ + 5x⁴+ x⁵) (1 + 5x¹⁰ + 10x²⁰+ ...)
= 10 x 10
= 100
probability = 100/ 4⁵
= 25 /256
Answered by
0
Answer:
25/256
total cases is equal to 256
probably cases is equal to 10×10
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