- FibonacciChick666

One card is selected from a standard deck of 52 cards. What is the probability that the card is either a diamond or a face card?

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- Nnesha

there are 13 face cards right :P o,O hmm

- FibonacciChick666

Looking for others' interpretations here. I thought it should be \(\frac{13}{52}+\frac{9}{52}=24/52=12/26=6/13\)

- FibonacciChick666

I'm reading the problem wrong somehow(or book has a type-o)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- FibonacciChick666

Yea, 13 face cards nosh

- Nnesha

and 13 diamonds lol ??? no idea abt cards :P

- amistre64

some rules define aces as faces

- FibonacciChick666

They say the answer is 13/52

- FibonacciChick666

Made that mistake ( I defined an ace as a face first time around)

- FibonacciChick666

but this is my second go

- FibonacciChick666

Shouldn't the probability be greater than just diamonds if we have an or and not an XOR?

- amistre64

if we remove all the face cards, and the diamonds, that is our ... outcome space?

- amistre64

10+12 = 22 cards that are favored, out of 52 total

- amistre64

1thru9 jkq
9+12 = 18 ... out of 52 :)

- FibonacciChick666

wait, only KQJ are face cards.... I think

- amistre64

XOR doesnt seem to translate well into english prose

- amistre64

its 11pm .. and i forgot how to add ... im going to bed :)

- FibonacciChick666

lol, I am a bit confused by your argument. Let me attempt to understand

- FibonacciChick666

we should have 30 cards not an option

- ganeshie8

|dw:1444012968296:dw|

- FibonacciChick666

3 suits, A-10 each

- FibonacciChick666

I added wrong above My conclusion should be 11/26

- Nnesha

told you then i deleted the comment
i thought this is impossible how fib can got it wrong lol

- FibonacciChick666

lol I can't add today, clearly

- FibonacciChick666

but still they are claiming 13/52 is the answer

- FibonacciChick666

@ganeshie8 Would you say 13/52 is incorrect?

- FibonacciChick666

I mean I think you proved it, but just want confirmation

- ganeshie8

A : diamonds
B : face cards
n(A) = 13
n(B) = 12
n(A \(\cap\) B) = 3
outcomes in favor = n(A \(\cup \) B)
= n(A) + n(B) - n(A \(\cap\) B)
= 13 + 12 - 3
= 22
total outcomes = 52
so p(A \(\cup\) B) = 22/52

- FibonacciChick666

here is their argument:
"The prob. that a card is either a diamond or a face card is determined as follows:
Let D be the event the card is a diamond. Let F be the event the card is a face card. Then, Pr[D}=13/52 and Pr[F]=12/52. We want to compute the probability of the event DUF
Pr[DUF]=Pr[D]+Pr[F]-Pr[D(intersect)F]=13/52+12/52-3/13=13/52

- FibonacciChick666

Ok, so book is def. wrong and I can't add... Great.
Thanks guys!

Looking for something else?

Not the answer you are looking for? Search for more explanations.