Let's assume that there are four difficult questions in the test paper of 10, and A and B draw one from each question in turn (don't put it back), so the probability of trying to draw difficult questi

Let's assume that there are four difficult questions in the test paper of 10, and A and B draw one from each question in turn (don't put it back), so the probability of trying to draw difficult questions has nothing to do with the order. The probability of winning the puzzle is 4/ 10 = 2/5.

After smoking, there are two situations that can be discussed:

( 1)

The person in front drew a difficult problem (the probability is 2/5). At this point, there are 3 difficult problems in the remaining 9 problems, so the probability is:

2/5 x 3/9=2/ 15

(2)

The person in front didn't get the question (the probability is 3/5). At this time, there are still 4 questions left in the 9 questions, so the probability is:

3/5 x 4/9=4/ 15

So the probability of drawing a difficult problem is 2/15+4/15 = 6/15 = 2/5.

So the chances of two people drawing difficult signatures have nothing to do with the order.