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...

Hello!

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.`

Thus `P(A_1A_2)=52/54*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

`52/54*51/53*50/52*49/51*48/50*2/49*1/48.`

Some numbers are reduced and we obtain

`2/(54*53)=1/(27*53)=1/1431 approx 0.0007.`  This is the answer.