Question

The number 1 through 10 are written in a separate slips of paper, and the slips are placed into a box. Then, 4 of the slips are drawn at random. What is the probability that the drawn slips are 1, 2, 3, and 4, in that order?

Answer 1

Given n(S)=10
let A be the event of drawing the slips 1, 2, 3, 4.
therefore
n(A)= 4
\[P(A)= \frac{n(A)}{n(S)}= \frac{4}{10}= \frac{2}{5}\]

@outbutharmless

Answer 2

Since you have to get a 1 on the first draw, and there are 10 slips, the probability
of getting a 1 is 1/10.
On the second draw, there are 9 slips left and you have to get a 2, so the probability.
is 1/9

The probability of getting a 1 and then a two is the product of the individual probabilities
1/10 x 1/9 =1/90.

Just keep going the same way for the third and fourth draw, and you get the answer...

Answer 3

1/10 x 1/9 x 1/8 x 1/7 =1/5040 = 0.0001984?