CAN SOMEONE SOLVE A QUESTION ON PROBABILITY?

I assume you mean if 1 is drawn, then 2 must be drawn next, then 3.

First we must look at the probability of the first card being drawn able to have this quality. It cannot be 9 or 8 because there is not a 10 card in the deck for 8,9, 10 to be drawn. Therefore there are 10-2=8 choices of numbers that would work from the first drawing. So 8/10 is the probability. Since every number only has one number one bigger than it, and one card was already removed from the deck, the probability of the next card being the correct one is 1/9 if the first card was right. By the same logic for the next card the probability is 1/8 if the second was right.

We multiply probabilities and get 8/10*1/9*1/8=1/90 which is the answer.

Note: The above poster is wrong. He should have done 10 Permutation 3 to find all possible permutations in the way that he did it. NOT 10 choose 3.

There are 8 possible ways that the three cards are selected in order.

123, 234, 345, 456, 567, 678, 789, 8910.

Then, 10 choose 3 is 120.

120/8=15.

Thus probability is 1/15 aka 6 2/3 %

ANS: 6.67% (3 sf)

You will certainly have picked 3 different numbers.

The number of ways these three numbers could be arranged is 3! = 6

Only one of these ways is in ascending order.

Hence

p = 1/6