# Find the probability that an ace will appear only in the fifth draw?

Five cards are drawn in succession and without replacement from an ordinary deck of playing cards. Find the probability that an ace will appear only in the fifth draw?

first, second , third and fourth draws must not be aces. $\frac{48}{52}$ x $\frac{47}{51}$ x $\frac{46}{50}$ x $\frac{45}{49}$ then you must draw an ace, there are 4 aces and 48 cards left. $\frac{4}{48}$. once computed the answer is 0.0598947271216

• I think this is too brief – QuIcKmAtHs Jan 25 '18 at 14:28

The probability that an ace will not appear in any of the first four draws is

$$\frac{48}{52} \cdot \frac{47}{51} \cdot \frac{46}{50} \cdot \frac{45}{49}$$

The probability of an ace appearing on the last draw is:

$$\frac{4}{48}$$

Just multiply the two results to get the final answer.

• "each draw is independent event". What do you mean? I cannot even recognize an event in a draw on its own. To "draw an ace" is an example of an event. Further the probabilities in your answer are - except for the first - conditional probabilities. E.g. $\frac4{48}$ is the probability that the fifth draw is an ace under condition that the first four draws were not. That does not correspond with independent. – drhab Jan 25 '18 at 14:26
• @drhab By independent event I mean we can consider each draw seprately and eventually multiplying the probabilites of each draw. The event is draw the first ace at the 5th draw. Which can be decomposed as drawing no ace in each of the first four draws and ace in the second. If we have that $A_i$ is the event of drawing an ace in the $i$-th draw then: $$P(\bar{A_1} \cap \bar{A_2} \cap \bar{A_3} \cap \bar{A_4} \cap A_5) = P(\bar{A_1})\cdot P(\bar{A_2})\cdot P(\bar{A_3})\cdot P(\bar{A_4})\cdot P(A_5)$$ – Stefan4024 Jan 25 '18 at 14:48
• If $A_2$ denotes the event of drawing an ace in the $2$-th draw then $P(\bar A_2)=\frac{48}{52}\neq\frac{47}{51}$. Here we have: $P(\bar A_1\cap\bar A_2\cap\bar A_3\cap\bar A_4\cap A_5)=P(\bar A_1)P(\bar A_2\mid\bar A_1)P(\bar A_3\mid\bar A_1\cap\bar A_2)P(\bar A_4\mid\bar A_1\cap\bar A_2\cap\bar A_3)P(A_5\mid\bar A_1\cap\bar A_2\cap\bar A_3\cap\bar A_4)$ – drhab Jan 25 '18 at 15:00