Adeck of cards contains 52 cards. they are divided into four suits: spades, diamonds, clubs and hearts. each suit has 13 cards: ace through 10, and three picture cards: jack, queen, and king. two suits are red in color: hearts and diamonds. two suits are black in color: clubs and spades.

use this information to compute the probabilities asked for below and leave them in fraction form. all events are in the context that three cards are dealt from a well-shuffled deck without replacement.

a. the first and second cards are both hearts.
b. the third card is an eight.
c. none of the three cards is an ace.

a)  13/52    and the second  12/51

b) Two solutions:

b.1 if we did not picked up an eight in the first two cards   4/50

b.2 there is an eight i the two previous    3/59

c ) (48/52)*(47/51)*(46/50)

Step-by-step explanation:

Condition: Cards are taken out without replacement

a) Probability of first card is heart

There are 52 cards and 4 suits with the same probability , so you can compute this probability in two ways

we have  13 heart cards and 52 cards  then probability of one heart card is  13/52   = 0.25

or you have 4 suits, to pick up one specific suit the probability is 1/4 = 0,25

Now we have a deck of 51 card with 12 hearts, the probability of take one heart is : 12/51

b) There are 4 eight (one for each suit )   P =  4/50 if neither the first nor the second card was an eight of heart, if in a) previous we had an eight, then this probability change to 3/50

c) The probability of the first card different from an ace is 48/52 , the probability of the second one different of an ace is 47/51 and for the thirsd card is 46/50. The probability of none of the three cards is an ace is

(48/52)*(47/51)*(46/50)

can you give me more detail on this! ?

step-by-step explanation:

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