10 signature, 3 difficult to sign, 2 signatures have been taken away.
There are three kinds of signatures: 1, and both signatures are difficult to sign. One of the two signatures is difficult to sign. It is not difficult to sign both of them.
The first case:
The probability of the selected two difficult signatures =C3.2 (subscript 3, superscript 2, and so on) //C 10.2.
The probability that the remaining eight are not difficult to sign is 7/8.
Therefore, the probability of drawing is not difficult to sign =C3.2/C 10.2 × 7/8.
The second case:
The two drawn lots are 1 difficult to sign = c 3.1* c 7.1/c10.2.
The probability that the remaining eight are not difficult to sign is 6/8.
Therefore, the probability of drawing is not difficult to sign = c 3.1* c 7.1/c10.2× 6/8.
The third situation:
The probability of whether the two draws are difficult to sign = c7.2/c 10.2.
The probability that the remaining eight are not difficult to sign is 5/8.
Therefore, the probability of drawing is not difficult to sign =C7.2/C 10.2 × 5/8.
The sum of the results of three situations is the answer you want:
c 3.2/c 10.2×7/8+c 3. 1 * c 7. 1/c 10.2×6/8+c 7.2/c 10.2×5/8
I made it myself, it should be right ~ ~ ~