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A turn consists of rolling a standard die and tossing a fair coin. The game is won when the die shows a 1 or a 6 and the coin shows heads. What is the probability the game will be won before the fourth turn? Express your answer as a common fraction.

Two distinct positive integers from 1 to 50 inclusive are chosen. Let the sum of the integers equal S and the product equal P. What is the probability that P+S is one less than a multiple of 5?

Guest Feb 6, 2019

#1**+1 **

2. CPhill's Answer:

Let one of the integers = a

Let the other integer = b

And 1 less than a multiple of 5 can be written as 5n - 1 where n is an integer ≥ 1

So....we have this equation

S + P = 5n - 1

(a + b ) + (ab) = 5n - 1 (1)

Rearranging (1), we have

a + ab + b = 5n -1

a + ab + b + 1 = 5n

a (b + 1) + 1 ( b + 1) = 5n

(a + 1)(b + 1) = 5n

Note that if "a" ends in a "4" or a "9' then (a + 1) is a multiple of 5, and no matter the integer value of b, the left side is always a multiple of 5. And the right side is definitely a multiple of 5

So...the "a's" that end in either 4 or 9 from 1-50 inclusive are :

4, 9 ,14, 19, 24, 29,34, 39, 44, 49

Notice that we can pair 4 with any of the other 49 integers and (1) will be true

Likewise, we can can pair 9 with any of 48 integers [we've already paired it with 4 ], and (1) will be true

And 19 can be paired with any of 47 other integers [ we've already paired it with 4 and 9 ] and (1) will be true

Continuing this reasoning with each successive number, we finally arrive at the fact that 49 can be paired with any of 40 other integers and (1) will be true

So....the number of pairs is just 49 + 48 + 47 + .....+ 41 + 40

And totalling these, we have that the number of pairs is just :

( 49 + 40) * 10 / 2 =

89 * 5 =

445 pairs a,b that make (1) true

And the number of possible pairs of a and b is C(50,2) = 1225

So, the probability that P + S is one less than a multiple of 5, is just

445 / 1225 =

89 / 245 ≈ 36.3 %

somebody Feb 6, 2019