Thread: probabilty

Suppose the two joker cards are left in a standard deck of cards. One of the jokers is red and the other is black. A single card is drawn from the deck of 54 cards but not returned to the deck, and then a second card is drawn. Determine the possibility of drawing...

a) any black card on both draws
b) any numbered card below ten on the first draw and the same number on a card on the second draw
c) the red joker or a red ace on either draw

Come on Johett, show us some work.
What have you done on any of these?
What is the probability the first card is black?

there is a) - f) for this question and I was able to get the first three however I can't get the answers for the last three... which are the above questions.


The 'original' questions for a)-c) are the following...I got the correct answers but I'm not sure if this is the right method to get the answer.

a)one of the jokers on the first draw and an ace on the second draw

2 x 4 = 4
54 53 1431

b)a numbered card of any suit on the first draw and the red joker on the second draw

36 x 1 = 2
54 53 159

c)a queen on both draws

4 x 3 = 2
54 53 477

Good for you. The method is correct.
Both black is the same as both queens.

If it's the same for d) as it is for c), are the answers suppose to be the same?

c)a queen on both draws
d) any black card on both draws


e) any numbered card below ten on the first draw and the same number on a card on the second draw

32 x 31 = 496
54 53 1431

2-9 ==>8 x 4
31: one more card is drawn

f) the red joker or a red ace on either draw

sorry if the letterings are a bit confusing

Originally Posted by johett

If it's the same for d) as it is for c), are the answers suppose to be the same?

Not the same answer. The same idea.

26 black cards out of 54 cards...

what would the second part be??

the answer is 13/53 so, I'm guessing, divide the total number of black cards in half to 13 and with a total of 53 (after a black card has been drawn)