You have a 54 card deck with 2 jokers, a normal deck has 52 cards. What is the probability of removing 5 cards from the deck and then the next 2 cards removed will be the 2 jokers?
The events considered in this problem are not independent. Therefore we have to use the formula of conditional probability:
`P(AB) = P(A)*P(B | A),`
where `A` and `B` are some events, `AB` is the event "`A` and `B`" and `B | A` is the event "`B` given event `A`".
We'll use this formula several times for all the seven elementary events, denote them `A_n.` The first event, `A_1,` is "the first card drawn is not a joker" (there are only 2 jokers, so to draw them on steps 6 and 7 all previous cards must not be a joker). Obviously `P(A_1)=52/54.`
Then we are interested in event `A_2,` "the second card is not a joker". The conditional probability `P(A_2 | A_1)` is also obvious: `54-1=53` cards remain and `52-1=51` are suitable, `P(A_2 | A_1)=51/53.`
The same way `P(A_1 A_2 A_3 A_4 A_5)=52/54*51/53*50/52*49/51*48/50.`
The events `A_6` and `A_7` are "the card drawn is a joker", and they give two more factors, `2/49` and `1/48.`
So the resulting probability is
Some numbers are reduced and we obtain
`2/(54*53)=1/(27*53)=1/1431 approx 0.0007.` This is the answer.