Find the probability that the third sword appears in the sixth extraction

Question

From a deck of playing cards, several cards are successively selected randomly and without replacement. Find the probability that the third sword appears in the sixth extraction.

Define

Let $A$: Event of obtaining two swords in the first five selected cards

$B$: The event of a sword in the sixth extraction.

The probability to find is: $P(A\bigcap B)$.

I know how to calculate $P(A)$ and $P(B)$ but I do not know how I can do to calculate $P(A\bigcap B)$, I think you have to use conditional probability, could someone help me please?

I believe that $P(A)=4\frac{\frac{4!}{2!2!}\frac{48!}{46!2!}}{\frac{52!}{50!}}$ and $P(B)=(\frac{48}{52})^5\frac{4}{52}$, am I right? Thank you very much.

Answer 1

Assuming by Sword Card you mean Jack, then

I know how to calculate $P(A)$ and $P(B)$ but I do not know how I can do to calculate $P(A\cap B)$, I think you have to use conditional probability, could someone help me please?

• $A\cap B$ is the event of obtaining two from four 'swords' and three from fourty-eight non-swords, when selecting any five from all fifty-two cards without replacement, and obtaining the one of the remaining two swords when drawing any one from the remaining fourty-seven cards.
• It is also the event that shuffling will place: two sword in the first five places, one in the next place, and one among the remaining fourty six places; when the shuffle selects places for the four swords in the deck of fifty-two cards.

I believe that $P(A)=4\frac{\frac{4!}{2!2!}\frac{48!}{46!2!}}{\frac{52!}{50!}}$ and $P(B)=(\frac{48}{52})^5\frac{4}{52}$, am I right?

No.   You are aiming at the right place, but are a little off the mark.

• $A$ is the event of obtaining two from four 'swords' and three from fourty-eight non-swords, when selecting any five from all fifty-two cards without replacement.

• It is also the event that shuffling will place two swords in the first five places and two in the remaining fourty-seven places, when the shuffle selects places for the four swords in the deck of fifty-two. $$\mathsf P(A)=\dfrac{\dbinom{4}{2}\dbinom{48}{3}}{\dbinom{52}5}=\cfrac{\cfrac{4!}{2!~2!}\cfrac{48!}{3!~45!}}{\cfrac{52!}{5!~47!}} =\cfrac{\cfrac{5!}{2!~3!}\cfrac{47!}{2!~45!}}{\cfrac{52!}{4!~48!}} =\dfrac{\dbinom 52\dbinom{47}2}{\dbinom{52}4}$$

• $B$ is the event of obtaining one from four swords when selecting one from fifty-two cards for the sixth draw.

$$\mathsf P(B)=\dfrac{\dbinom{4}{1}}{\dbinom{52}{1}}$$

• Now, $A$ and $B$ are not independent, so you will have to use conditional probability, or calculate using the suggestion above.