mathmath333
  • mathmath333
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.
Mathematics
schrodinger
  • schrodinger
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{ 6 positive number are taken at random and multiplied together.}\hspace{.33em}\\~\\ & \normalsize \text{ Then what is the probability that the product ends in an odd }\hspace{.33em}\\~\\ & \normalsize \text{ digit other than 5. }\hspace{.33em}\\~\\ \end{align}}\)
mathmate
  • 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)?
mathmath333
  • mathmath333
2/5 ?

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More answers

mathmate
  • mathmate
It's 2/5 if you only pick one single number. Use the multiplication rule for 6 numbers that are multiplied together.
mathmath333
  • mathmath333
I still cant get ur hint
dan815
  • dan815
=]
mathmath333
  • mathmath333
i m depressed
dan815
  • dan815
hahahaha
dan815
  • dan815
im actually laughing irl rn
dan815
  • dan815
dang mathmate wrong something huge, can i just do it over
dan815
  • dan815
wrote*
dan815
  • dan815
6 numbers multiplied together no even, no 5 on the end
mathmath333
  • mathmath333
is it =2/5 *6
dan815
  • dan815
so no even numbers picked, and no 5 picked as that will gurantee a 5 or even number ending
dan815
  • dan815
the multiplication of all odd numbers is always off so u are safe there
dan815
  • dan815
odd*
mathmath333
  • mathmath333
answer was (0.4)^6
mathmath333
  • mathmath333
actuallyi m poor in english , cant totaly understand matemate paragraph
dan815
  • 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
dan815
  • dan815
the explaination is still same as above
dan815
  • dan815
what language do you speak?
mathmath333
  • mathmath333
rural marathi/hindi
mathmath333
  • mathmath333
हॅलो दान
mathmath333
  • mathmath333
https://www.youtube.com/watch?v=GHEX2RrY-cY
dan815
  • dan815
okay okay
dan815
  • dan815
do u want to start over?
mathmath333
  • mathmath333
how come \( (10^6/6!)\)
dan815
  • dan815
thats all the possible 6 digit products there are
dan815
  • dan815
this is another way to think about it uhh
dan815
  • dan815
u know we can pick only 4 digits out of the 10 digits
dan815
  • dan815
so (4/10)^6
mathmath333
  • mathmath333
but why ^{6}
dan815
  • dan815
for 6 digits
dan815
  • dan815
|dw:1444592494691:dw|
dan815
  • dan815
tbh we shudnt even be thinking of just 10 digits lol
dan815
  • dan815
its the fact that 4/10 of an infinite possible positive numbers
mathmath333
  • mathmath333
ok

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