Help need for repeat task(solved)

Question

import random
A = [10,20,30,40,50,60]
random.shuffle(A)
X = (A[0] / 10)
if X > 2:
print("Ok" ,X)
else:
print("Not Ok", X)
# How i get always > 2.0 in this code
Dont want to print Not Ok
#EDIT
SOLVED
#It will give always print Ok
import random
A = [10,20,30,40,50,60]
random.shuffle(A)
X = (A[0] / 10)
if X >= 3:
print("Ok" ,X)
else:
print("Not Ok and Retrying....")
X = (A[1] / 10)
if X >= 3:
print("Ok" ,X)
else:
print("Not Ok and Retrying....")
X = (A[2] / 10)
if X >= 3:
print("Ok" ,X)

Answer 1

Why? May I what exactly you are trying here?
Edit: ғᴜɴ 3x
import random
A = [10,20,30,40,50,60]
while True:
random.shuffle(A)
X = (A[0] / 10)
if X > 2:
print("Ok" ,X)
break
#2nd way
import random
A = [10,20,30,40,50,60]
random.shuffle(A)
i=0
while True :
X = (A[i] / 10)
if X >= 3:
print("Ok" ,X)
break
else:
print("Not Ok and Retrying....")
i++

Answer 2

You can do it more ways.. ғᴜɴ 3x
Put X=A[0] only.. Instead A[0]/10 //OK?
But still it is not clear...
Edit :
Is you want to print otherthan 10 and 20 from list....?

Answer 3

Want to give me ans only 2 or up

Answer 4

After random.suffle(A) please add print(A).
Your code is working but you have a probability of 1/6 to output Not Ok. Do it ten times
import random
A = [10,20,30,40,50,60]
for i in range(10):
random.shuffle(A)
print(A)
X = (A[0]/10)
if X > 2:
print("Ok" ,X)
else:
print("Not Ok", X)

Answer 5

Jayakrishna🇮🇳
Thanks A Lot Bro 👌👏👏
I like first one while loop 👌