Here is the question:
An urn contains balls numbered 1 through 5. 6 balls are selected, one at at time, and with replacement. What is the probability that each ball is selected at least once?
My thoughts:
Let $A$ be the event where each ball is selected at least once. Since 6 balls must be selected but there are only 5 choices for each draw, that means only one of the numbers can repeat. Since each number must appear once, I would have $5!*5$ in the numerator, where the first 5 balls must not have the same number but the last ball can be any of the 5 numbers. The possible number of outcomes is $5^6$. I then set up the solution like this: $$P(A)=\frac{5!*5}{5^6}$$
However, this is incorrect and I am having trouble understanding why. Any advice would be appreciated. Thank you.
Best Answer
As each time a ball is drawn with replacement, the probability of a specific number showing up is
$P_1 = \frac{1}{5}$
As each numbered ball must draw at least once, one ball will draw twice.
Number of ways of choosing a numbered ball that is drawn twice = $^5C_1$
Permutations of $6$ numbered balls (with one number appearing twice) $= \frac{6!}{2!}$
So, the desired probability $ \displaystyle = \frac{^5C_1 \times 6!}{2!} \times (P_1)^6 = \frac{3 \times 4!}{5^4}$
Alternatively -
Number of ways of drawing $5$ numbered balls in $6$ draws with the first ball drawing twice (say repeat draw is in the second drawing) -
$ = \, ^5C_1 \times \, ^1C_1 \times \, ^4C_1 \times \, ^3C_1 \times \, ^2C_1 \times \, ^1C_1 = 5!$
As the first ball can repeat once in any of the $5$ drawings $(2 - 6)$ -
$C(1) = 5 \times 5!$
Number of ways to draw with the second ball drawing twice $C(2) = 4 \times 5!$
Total number of desired ways $C = \sum \limits_{i = 1}^{5} C(i) = 5! \times (5+4+3+2+1)$
Total number of ways to draw balls without restrictions $= 5^6$
So desired probability $ \displaystyle = \frac {15 \times 5!}{5^6} = \frac {3 \times 4!}{5^4}$