Question

This is talking about disjointed/ mutually exclusive. (They can't happened at the same time.)

One card is randomly selected from a stranded deck. Compute the probability
A. Of randomly selecting a two or three
B. Of randomly selecting a two or club from a deck of cards.
C.not selecting a heart or a face card.

Answer 1

your thoughts?

Answer 2

i got no idea what a stranded deck is ...

Answer 3

standard maybe?

Answer 4

yes, sorry.

Answer 5

2 or a 3 ... how many cards in the deck have a 2 or a 3 on them?

Answer 6

the mathical formula would be:

P(A) + P(B) - P(AnB)

Answer 7

there are 4 2's and 4 3's

Answer 8

good, so 8 cards that we can pull that are either a 2 or a 3

8 out of 52 right?

P(2) = 4/52
P(3) = 4/52
P(2 n 3) = 0/52

so the formula works fine as well

Answer 9

if we remove all the 2s and all the clubs from the deck, how many cards do we have?

Answer 10

just to butt in, it is "\(\text{disjoint"}\) not \(\text{disjoint}\color{red}{ed}\)

Answer 11

well, maybe a stranded deck of cards IS disjointed lol

Answer 12

what does P(2 n 3) = 0/52 suppose to mean?

And yes lol i know, sorry for the typos lol

Answer 13

the probability that a card is both a 2 and a 3 is, 0 out of 52 cards.

Answer 14

ook, i understand. it just looked weird lol

Answer 15

math is weird, yes

Answer 16

p(2 or 3) = 4/52 + 4/52

Answer 17

yes ... but for completness, we would want to excluded double counting so we remove the set that is common to both 2 and 3 ... otherwise we count them twice.

Answer 18

8/52 = 4/26 = 2/13 =0.1539

is that right?

Answer 19

oh, what? sooooo..

Answer 20

consider the venn:


n(A or B) = n(A) + n(B)
= (1+2)+(2+3)

we have something counted twice, we have to remove it ... n(AnB)

n(A or B) = n(A) + n(B) - n(AnB)
= (1+2)+(2+3) - (2)

Answer 21

how would you know something is counted twice?

Answer 22

the other questions make this more obvious ...

Answer 23

how many cards have a 2 on them?

how many cards are clubs?

if we count all the 2s with all the clubs .... are we overcounting something?

Answer 24

yes

Answer 25

this is consider a dependent situation, right?

Answer 26

dependence is a specific concept in probability ... it is easily confused with mutually exclusive as a word, but it is defined as something entirely different.

if the probability of an event is the same for any given event that it occurs in .... then the probability of the event does not depend on a given situation.

this does not mean that the events are or are not mutually exclusive ... its just that the probability works out so that it is the same regardless.

Answer 27

spose that 1/4 of the m&ms in a bowl are green.

spose that after eating 12 m&ms, we determine that 1/4 of the remaining m&ms are still green.

in this instance, the probability remains the same regardless of which event we observe. so we say conclude that it simply doesnt matter which events probability we use.

Answer 28

P(green)= P(green, given 12 m&ms are missing)

1/4 = 1/4

the probability of a green m&m is the same regardless of the event we choose ... the probability is independant of the event that you calculate it for.

Answer 29

ok, i think i get what you're saying.

Answer 30

say 1/3 of the class has blond hair

and given that we pull out all the girls, we find that 1/3 of the girls have blond hair.

the probability of picking a blond is the same for each event, the probability does not depend on choosing from the class or from the subset.

Answer 31

ok.

Answer 32

Its starting to make sense to me

Answer 33

just drilling in that words used in math have more precise definition then assuming they are just english words

Answer 34

Mathical terminology should not have 'synonyms'

English usage and math usage should be compartmentalized in your head :)

Answer 35

I actually agree

Answer 36

so

number of 2s in a deck: 2,2,2,2

number of clubs in a deck: A,2,3,4,5,6,7,8,9,J,Q,K

4+13 is not a good count is it? something has been counted more than once in this case.

a single card is both a 2 and a club ... we counted it in both sets, we need to remove the duplication.

n(2) = 4
n(clubs)=13
n(2 n clubs)=1

4+13-1 is a proper count for the desired outcomes right?

Answer 37

yes, i agree

Answer 38

so, it would be 4/52 + 13/52 = 17/52 - 1/52 = 16/52

Answer 39

correct

Of randomly selecting a two or club from a deck of cards.

(4+13-1) out of (52) is our results

Answer 40

now the last questions is:

not selecting a heart or a face card

how many hearts? how many face cards? how many that are both heart and face?

Answer 41

c. is of selecting, not, not selecting. that was a typo
real question: of selecting a heart or a face card

Answer 42

so, in this situation, i would need to find how many hearts and how many faces, right?