SOLUTION: A pack of cards consist of 4 types of cards, diamond, hearts, spades , clubs. Each pack contains 13 diamonds, 13 hearts, 13 spades and 13 clubs in the total of 52 cards. There are

Algebra ->  Probability-and-statistics -> SOLUTION: A pack of cards consist of 4 types of cards, diamond, hearts, spades , clubs. Each pack contains 13 diamonds, 13 hearts, 13 spades and 13 clubs in the total of 52 cards. There are       Log On


   



Question 752626: A pack of cards consist of 4 types of cards, diamond, hearts, spades , clubs. Each pack contains 13 diamonds, 13 hearts, 13 spades and 13 clubs in the total of 52 cards. There are 26 red and 26 black cards, diamonds and hearts are the red cards and spades and clubs are the black cards. In any one of the 4 types there are 9 numbered cards, numbered 2,3,4,5,6,7,8,9, a jack and a quen, a king and a ace, if a card is drawn from the pack it can be a combination of the above.
given the infor, what is the probability that
A) a red card will be drwn
B) a spade card will be drawn
C) a Ace will be drawn
D) A jack of hearts will be drawn

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A pack of cards consist of 4 types of cards, diamond, hearts, spades , clubs. Each pack contains 13 diamonds, 13 hearts, 13 spades and 13 clubs in the total of 52 cards. There are 26 red and 26 black cards, diamonds and hearts are the red cards and spades and clubs are the black cards. In any one of the 4 types there are 9 numbered cards, numbered 2,3,4,5,6,7,8,9, a jack and a quen, a king and a ace, if a card is drawn from the pack it can be a combination of the above.
given the infor, what is the probability that
A) a red card will be drawn::: 26/52 = 1/2
----------------------------------------------
B) a spade card will be drawn::: 13/52 = 1/4
------------------------------------------------
C) a Ace will be drawn:::: 4/52 = 1/13
------------------------------------------------
D) A jack of hearts will be drawn:::: 1/52
================================================
Cheers,
Stan H.