Question

PLEASSEEE HELLLPPP! I will give medal :)

Two cards are drawn from a standard deck of cards.

Part A: If they are drawn with replacement, what is the probability that both cards are 10s?

Part B: If they are drawn without replacement, what is the probability that the first card is a club and the second card is a diamond?

Part C: Which of the two scenarios in Part A or Part B represents dependent events? Explain.

Answer 1

Since there is no replacement, the probability of the second draw depends on the first outcome. Let me illustrate on the part a.

a. There are 13 diamonds in the deck (just as any other suit). Since there are 52 cards, the probability of getting a diamond in the first draw is 13/52. After the first card is drawn, there are just 12 diamonds left. So, the probability of drawing the diamond now is 12/51 (remember, there is no replacement, so there are just 51 cards left after the first card is drawn!). The total probability is the product of two probabilities: the probability to draw diamond from original deck, which is 13/52 times the probability to draw a diamond from the remaining 51 cards, provided the first card drawing is diamond, which is 12/51.
So, p=13/52*12/51=1/4*4/17=1/17; Answer to a. part is 1/17.

Now you shall be able to answer part b. by analogy. I will give just the answer: 19/34. Let us consider part c now.

There are 13 units in the deck, with 4 cards of the same unit per each unit. So, drawing a given unit has a probability of 4/52=1/13. Once a card of a certain unit is drawn, there are just 3 cards of the same unit left, so the probability of drawing the second of the same unit is 3/51=1/17. The product of two probabilities is the total probability to draw two cards of the same given unit, that is 1/(13*17)=1/221. Since there are 13 units, this needs to be multiplied by the number of units, that is 13. Thus the answer is 1/221*13=1/17.

Now to part d. If the first card is Club, then there are 12 Clubs left and overall 51 cards left. Probability to draw a Club second time is 12/51=4/17.

Part e. Let us consider the probability that they are of the same suit. In part a. we determined the probability to have two diamonds. It is 1/17. So is the probability to get two cards of any other suit. Thus the total probability to get two cards of the same suit is 4*1/17=4/17. The probability that two cards are not of the same suit is just 1-4/17=13/17. Answer: 13/17.

Part f. If the first card is a Club, then the second MUST be NOT Club. There are 12 Clubs left and total 51 cards left. Thus there are 51-12=39 cards, that are NOT Clubs, left. Thus the probability to draw the second card, which is not a Club, is 39/51=13/17. Answer: 13/17.

Answer 2

Can you help me with some more? @trevor12