Given that x andy are any two consecutive positive numbers such that neither of them are divisible by 4. What is the probability of finding x +y +1 to be divisible by 4?
Answers
Answer:
Let two consecutive integers are x and x+1.
Since the two integers would be odd and even or even or odd, the remainder would be odd and even or even and odd. Thus sum would always be odd except when the 1st integer is an even integer and 2nd is a multiple of 5.
Example: For 10 and 11,
For 11 and 12,
....
For 14 and 15,
(1) The sum of the remainders is even.
For 14 and 15,
Given: x and y are any two consecutive positive numbers such that neither of them are divisible by 4.
To find: Probability of x+y+1 to be divisible by 4
Solution: Let n be any natural number.
Now, for a number to be divisible by 4, it must be in the form of 4n.
For example- 12 can be written as 4×3 where n=3
20 can be written as 4×5 where n=5
Now, two consecutive positive numbers such that neither is divisible by 4 can be
1- 4n+1 and 4n+2 (For example- 13 and 14)
2- 4n+2 and 4n+3( For example- 14 and 15)
Now, after 4n+3 the next number 4n+4 is divisible by 4 as it can be written as 4(n+1) where n+1 is a natural number.
Taking the first case:
x= 4n+1
y= 4n+2
x+y+1 = 4n+1+4n+2+1 = 8n+4
which can be written as 4(2n+1) where 2n+1 is also a natural number and hence, divisible by 4.
Taking the second case:
x= 4n+2
y= 4n+3
x+y+1= 4n+2+4n+3+1= 8n+6
Here, 8n+6 can not be written in the form of 4n and therefore, it is not divisible by 4.
Therefore, out of the two possible cases, only one case is such that x+y+1 is divisible by 4.
Probability of x+y+1 divisible by 4
= Cases where x+y+1 divisible by 4 / Total cases
= 1/2
= 0.5
Therefore, the probability of x+y+1 divisible by 4 is 0.5 .