Question

Luca keeps picking playing cards out of a standard deck of 52 cards, hoping he will draw an ace. There are four aces in the deck. After looking at each card, he places it back in the deck. What is the probability that Luca will draw an ace in the first nine attempts?

Answer 1

When you draw a card, the possibility of drawing an ace is 4/52 cards, given that the 4 is the number of aces that are in the deck. If you simplify this, it would be 1/13. The probability of drawing two aces is around 0.45%. Hope I helped! If not, then at least I tried, right?

Answer 2

Answer with explanation:

Number of cards in the deck=52

Number of Aces in the deck of cards =4

Probability of an event

`=\frac{\text{Total favorable outcome}}{\text{Total Possible outcome}}`

Probability of drawing an ace

`=\frac{_(1)^(4)\textrm{C}}{_(1)^(52)\textrm{C}}\\\\=(4)/(52)\\\\=(1)/(13)`

Probability of not drawing an ace

`=1-(1)/(13)\\\\=(12)/(13)`

Probability that Luca will draw an ace in the first nine attempts

`=S+FS+FFS+FFFS+FFFFS+FFFFFS+FFFFFFS+FFFFFFFS+FFFFFFFFS\\\\=(1)/(13)+(12)/(13)*(1)/(13)+(12)/(13)*(12)/(13)*(1)/(13)+(12)/(13)*(12)/(13)*(12)/(13)*(1)/(13)+(12)/(13)*(12)/(13)*(12)/(13)*(12)/(13)*(1)/(13)+(12)/(13)*(12)/(13)*(12)/(13)*(12)/(13)*(12)/(13)*(1)/(13)+(12)/(13)*(12)/(13)*(12)/(13)*(12)/(13)*(12)/(13)*(12)/(13)*(1)/(13)+(12)/(13)*(12)/(13)*(12)/(13)*(12)/(13)*(12)/(13)*(12)/(13)*(12)/(13)*(1)/(13)+(12)/(13)*(12)/(13)*(12)/(13)*(12)/(13)*(12)/(13)*(12)/(13)*(12)/(13)*(12)/(13)*(1)/(13)\\\\=(1)/(13)*[1+(12)/(13)+((12)/(13))^2+((12)/(13))^3+((12)/(13))^4+((12)/(13))^5+((12)/(13))^6+((12)/(13))^7+((12)/(13))^8]\\\\=(1)/(13)*[(1-((12)/(13))^9)/(1-(12)/(13))]\\\\=(1)/(13)*[(1-((12)/(13))^9)/((1)/(13))]\\\\=1-((12)/(13))^9\\\\=1-0.4866\\\\=0.5134{\text{approx}}`