Question

A standard poker deck of cards contains 52 cards, of which four are tens. Suppose two cards are drawn sequentially, so that one random circumstance is the result of the first card drawn and the second random circumstance is the result of the second card drawn. Find the probability that the first card is a ten and the second card is not a ten.

Answer 1

The probability of only one 10 being drawn in two draws is given by
\[P(one\ 10\ out\ of\ 2\ draws)=\frac{4C1\times 48C1}{52C2}\]
There are two permutations of the possible outcomes of the 2 draws, only the permutation with a 10 drawn first is required.
Therefore the result of the above calculation must be divided by 2 to find the required probability.

Answer 2

Since the cards are drawn sequentially, the probability of drawing a 10 on the first draw is obviously 4 out of 52. Assuming that a 10 is drawn, and knowing that there are now only 51 cards left in the deck, the probability that a card other than a 10 will be drawn on the second go is 48 out of 51.

Answer 3

The method that I have shown does not have to assume that a 10 is drawn on the first draw and could be said to be more rigorous than the method posted by @EulersEquation. Both methods give the same answer.

Answer 4

i think that the @EulersEquation method is easier to understand

Answer 5

Actually, I gave an incomplete reply. But I was just showing the reasoning that leads to the solution.

Answer 6

@EulersEquation so u just need to multiply both probabilities right?

Answer 7

yes.

Answer 8

reduce first, and save yourself the trouble