Hi,
Can we do:
Total number of ways= 5!/(3!x2!) = choose 2 among 5 of the number unordered = 10
Only 1 combination unordered not favorable (2;4)
Not favorable probability = 1 /10
Favorable case = 1 - 1/10 = 9/10
?
Can we do:
Total number of ways= 5!/(3!x2!) = choose 2 among 5 of the number unordered = 10
Only 1 combination unordered not favorable (2;4)
Not favorable probability = 1 /10
Favorable case = 1 - 1/10 = 9/10
?
