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Since n! is the product of all the numbers from 1 through n and (n+1)! is everything in n! multiplied by n+1, the quotient is n+1 ■
m = n/(n-1)
N is 98 and M is 1
Let the integers be m and n.∴ By the condition given in the question,m + n - mn = 72∴ m + n - mn = 71 + 1∴ m + n - mn - 1 = 71∴ (m - 1)(1 - n) = 71∴ (m - 1)(1 - n) = 1 × 71 = 71 × 1 = (-1) × (-71) = (-71) × (-1)Case 1:m - 1 = 1 and 1 - n = 71∴ (m, n) = (2, -70)Case 2:m - 1 = 71 and 1 - n = 1∴ (m , n) = (72, 0)This case is inadmissible as m, n are non zero integers.Case 3:m - 1 = -1 and 1 - n = -71∴ (m , n) = (0, 72)This case is inadmissible as m, n are non zero integers.Case 4:m - 1 = -71 and 1 - n = -1∴ (m, n) = (-70, 2)But as the given expression is symmetric, therefore, (2, -70) and (-70, 2) cannot be considered two different pairs.∴ We get only one solution.
Proof: P{T>n+m/T>n}=P{T>n+m,T>n}/P{T>n} (Bayes theorem) =P{T>n+m}/P{T>n} =((1-p)^(n+m))/(1-p)^n = (1-p)^(n+m-n) = (1-p)^m (1-p)^m = {T>m} So T>m has the same probability as T>m+n given that T>n, which means it doesn't care (or don't remember) that n phases had passed.