This is a two part question:
The digits $1, 3, 3, 3, 4, 5$ are arranged to form a number with no repetitions, find the probability that the number will be:
a) Greater than 200,000
b) Odd
For a), I understand that the probability will be the number of permutations, where a number greater than 200,000, is divided by the total arrangements which is $6!\over 3!$ due to the repetition of 3. But when calculating the permutations for $n>200,000$ there is apparently $5!\over 3!$ ways of doing so according to official answers and so there is a $5!\over 3!$÷ $6!\over 3!$=$1\over 6$ probability, but $5!\over 3!$ only accounts for the last 5 digits of $n$ because the first digit may be $3, 3, 3, 4, 5$, so there should be $5(5!)\over 3!$ ways of arranging $n>200,000$, why am I wrong?
Secondly, b) states the probability of obtaining an odd number is $1-$($5!\over 3!$÷ $6!\over 3!$)=$5\over 6$, but it seems that it is also the probability of arranging $n≤200,000$ because it is the complementary probability. So this is an assumption that all permutations $n≤200,000$ are odd but obviously there can be $n=133354 , 133534$ and other permutations ending in four, so how is this calculation correct?
The probability that the number is greater than $200,000$ equls to the probability that the first digit is not $1$, which is $1 - 1 / 6 = 5 / 6$.
The probability that the number is odd equals to the probability that the last digit is not $4$, which is $1 - 1 / 6 = 5 / 6$.