Question

One card is drawn from a well shuffled deck of 52 cards. If each outcome
is equally likely, calculate the probability that the card will be
(i) a diamond
(ii)
not an ace
(iii) a
black card (i.e., a club or, a spade)
(iv)
not a diamond
(v) not a black card.

Answer 1

@oldrin.bataku

Answer 2

How many diamonds are there in a standard deck? How many aces? How many black cards?

Answer 3

When a card is drawn from a well shuffled deck of 52 cards, the number of
possible outcomes is 52

Answer 4

There are 13 diamond cards and 52 total. Our probability of picking a diamond card is then just \(13/52\).

Answer 5

Similarly, there are 4 aces (one per suit). So the number of cards that AREN'T aces is then merely 52 - 4 = 48, right? So the probability of picking a card that is not an ace is \(48/52\).

Answer 6

ok

Answer 7

Note there are 13 club cards as well as 13 spade cards, so in total we have 26 black cards. Thus the probability of picking a black card is just \(26/52=1/2\).

Answer 8

ok....

Answer 9

Note that we may reduce our earlier probabilities to \(13/52=1/4\) and \(48/52=12/13\).

Answer 10

ok

Answer 11

Since there are 13 diamond cards, we know there are 52 - 13 = 39 non-diamond cards. Thus the probability of picking a non-diamond card is \(39/52=3/4\).

Similarly, since there are 26 black cards, there are 52 - 26 = 26 non-black (red) cards. Thus the probability of picking a non-black card is \(26/52=1/2\).

Note that the sums of the probabilities for diamond and non-diamond cards sum to \(1=100\%\); this should make intuitive sense to you. *Every* card is either a diamond or a non-diamond card, so the probability of picking a card that is either diamond or non-diamond is the probability of picking a card at all - \(52/52=1\)! This means we didn't even need to count explicitly for these past two problems; consider that since the probability of picking a diamond card is \(1/4\), the probability of picking a non-diamond card is then just \(1-1/4=3/4\). Similarly, since picking a black card occurs with probability \(1/2\), picking a non-black card occurs with probability \(1-1/2=1/2\).

Answer 12

In fact, in these problems, counting has only slowed us down. If you understand the composition of a standard 52-card deck, the answers are clear. The deck is composed of an equal number of cards of each suit -- diamonds, hearts, clubs, and spades. The probability of picking a diamond card is then just \(1/4\), as is the probability for picking a hearts, clubs, or spades card.

Answer 13

Similarly, each suit has only one ace card out of their total 13 ranks. Thus the probability of picking an ace out of the entire thing is just \(1/13\), and the probability of picking a non-ace card is just \(1-1/13=12/13\). :-)

Answer 14

Notice that our suits are divided nicely not only into four distinct categories but also into colors, red (hearts, diamonds) and black (clubs, spades). So half of our total cards will be red or black, and thus the probability of picking a black card is just \(1/2\) -- as is that of picking a non-black (i.e. red) card.