Probability question

Question

There are 52 cards in a deck. 26 are red, and 26 are black. The 52 cards make up four suits (hearts, diamonds, spades, clubs). There are 13 of each suit (ace-10, jack, queen, king). Essentially it is a fair deck of cards.

f) What is the probability of drawing the 10 of clubs or a king, and then a spade?

g) What is the probability of drawing a 7 or a heart, and then a 10 of hearts or a queen?

Answer 1

There are two possibilities here, are the cards being replaced after drawing or not? The key difference is that if the cards are replaced, there are always 52 cards for each draw. If there is no replacement, then the number of cards to draw from decreases.

First, with replacement, on the first draw there are five cards that satisfy the conditions (10 of clubs and four kings) so that probability is 5 / 52. For the second draw there are 13 spades in the deck, so that probability is 13 / 52, or 1 /4. Now multiply the two together since these are independent events that both must occur. That gives (5 / 52) x (1 / 4), or 5 / 208.

If there is no replacement, then the second probability changes to 13 / 51. Now the overall probability becomes (5 / 52) x (13 / 51) = 65 / 2652 = 5 / 204.

For the second problem, with replacement, The there are 16 cards that satisfy the condition, There are 4 cards valued 7, and 13 hearts, but one of those is the 7 of hearts and cannot be counted twice. That probability is 16 / 52, or 4 / 13. On the second draw there are 5 cards that satisfy the condition, the 10 of hearts and four queens, so the probabilkity is 5 / 52. Now multiply as you did in the prior problem and the probability is (4 / 13) x (5 / 52) = 20 / 676 = 5 / 169.

Without replacement the deck is reduced to 51 cards for the second draw and the second probabilkity becomes 5 / 51. Multi[plying gives the overall probability as (4 / 13) x (5 / 51) = 20 / 663.

Hope this helps.