Tough probability question: What is the probability that a number formed using each of the digits 0-9 once is an even number over 6 billion?
1/2
no, 1/5
Good, show me how?
it can start with a zero, and the answer is 1/5
Nevermind then.
lol
1/5
over 6 billion or greater than or equal to 6 billion?
It cannot be = to 6 billion
Zark it doesn't matter since each digit is used once. No way it can be exactly 6 billion.
doesnt matter really because if the first number is a 5 its going to be less and if its 6 its going to be greater
ok
we need to make a 10 digit no. from 0-9 without repitition... total no. of ways are 10! now we need to make cases: CASE I : no. ends with a zero so total ways of making a no. over 6 billion are like at first position only 4 nos. are allowed i.e 6,7,8,9 and last is zero, rest arranged by 8! so total such nos. are 4X8! CASE II: no.ends with 6 or 8 for last place we have 2 choices for first place we have only 3 choices since one of 6 or 8 has been consumed at the last place rest arrange as 8! so total of this case = 6X8! CASE III: no. ends with 2 or 4 last place; 2 choices first place; 4 choices rest arrange as 8! total for this case = 8X8! now total favourable cases = 8X8! + 6X8! + 4X8! = 18X8! probability = (18X8!)/10! = 1/5
then 1/5
\[\frac{2\cdot 5\cdot 8!+2\cdot 4\cdot 8!}{10!}\]
man i wish satellite was here, zarkon can you explain that
didn't you get that....????
yes but you're way is long, satellite can do these like zarkon did there which is useful for me on quick, timed tests
the first digit can be 6,7,8,9 look at 6 and 8 first ...since the ones digit has to be even we have only 4 choices remaining. now pick the 8 remaining numbers for the middle..thus 2*4*8!
if we use 7 or 9 for the first digit we have 5 choices for the ones digit...
why do you use 8!?
8! ways to arrange 8 numbers in a row
okay
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