## mathmath333 one year ago 6 positive number are taken at random and multiplied together. Then what is the probability that the product ends in an odd digit other than 5.

1. mathmath333

\large \color{black}{\begin{align} & \normalsize \text{ 6 positive number are taken at random and multiplied together.}\hspace{.33em}\\~\\ & \normalsize \text{ Then what is the probability that the product ends in an odd }\hspace{.33em}\\~\\ & \normalsize \text{ digit other than 5. }\hspace{.33em}\\~\\ \end{align}}

2. mathmate

I assume "random" means that the probability of the last digit is uniformly distributed over the range 0-9. The resulting last digit depends only on the last digit of the six numbers, so without loss of generality (WLOG), we will consider only the product of six "random" digits. The resulting digit will be odd if and only if all six digits are odd. So what is the probability of getting 6 odd digits out of a uniform distribution? Out of the set {0,1,2,3,4,5,6,7,8,9}, there are 5 odd digits. So the probability of randomly selecting an odd digit is 5/10=1/2, resulting in an odd number. However, we know that 5 multiplied by any odd number will have 5 as a terminating digit, so we need to exclude 5 in ALL of the six numbers. That leaves us with 4, A={1,3,7,9} to choose from (out of 10). The probability of choosing any element of A from 10 digits is 4/10=2/5. The experiment is a 6 step experiment, each one independent of each other, so the multiplication rule applies. Can you figure out the probability of the final result (terminating with an odd digit excluding 5)?

3. mathmath333

2/5 ?

4. mathmate

It's 2/5 if you only pick one single number. Use the multiplication rule for 6 numbers that are multiplied together.

5. mathmath333

I still cant get ur hint

6. dan815

=]

7. mathmath333

i m depressed

8. dan815

hahahaha

9. dan815

im actually laughing irl rn

10. dan815

dang mathmate wrong something huge, can i just do it over

11. dan815

wrote*

12. dan815

6 numbers multiplied together no even, no 5 on the end

13. mathmath333

is it =2/5 *6

14. dan815

so no even numbers picked, and no 5 picked as that will gurantee a 5 or even number ending

15. dan815

the multiplication of all odd numbers is always off so u are safe there

16. dan815

odd*

17. mathmath333

18. mathmath333

actuallyi m poor in english , cant totaly understand matemate paragraph

19. dan815

hence we have 4 choices for each of the places so 4^6/6!, as we dont care about the order u pick the 6 numbers so (4^6/6!) / (10^6/6!) = 0.4^6

20. dan815

the explaination is still same as above

21. dan815

what language do you speak?

22. mathmath333

rural marathi/hindi

23. mathmath333

हॅलो दान

24. mathmath333
25. dan815

okay okay

26. dan815

do u want to start over?

27. mathmath333

how come $$(10^6/6!)$$

28. dan815

thats all the possible 6 digit products there are

29. dan815

this is another way to think about it uhh

30. dan815

u know we can pick only 4 digits out of the 10 digits

31. dan815

so (4/10)^6

32. mathmath333

but why ^{6}

33. dan815

for 6 digits

34. dan815

|dw:1444592494691:dw|

35. dan815

tbh we shudnt even be thinking of just 10 digits lol

36. dan815

its the fact that 4/10 of an infinite possible positive numbers

37. mathmath333

ok