Question

will medal!!!!
What is the probability of drawing three queens from a standard deck of cards, given that the first card drawn was a queen? Assume that the cards are not replaced.
I'm not sure how to set up this problem

Answer 1

Is your question asking what is the probability of drawing three queens on your next three turns?

Answer 2

yes its supposed to be a conditional probability question

Answer 3

Ok so there are 52 cards in a deck, of those 4 queens

Answer 4

youve already picked one queen, so there are now 51 cards and 3 queens

Answer 5

what is the chance that you will pick a queen on your next turn?

Answer 6

I tried to set it up like 4/52 times 3/51times 3/50 but i didn't get an answer that matched the answer choices

Answer 7

2/50 not 3/50

Answer 8

youre actually really close

Answer 9

since you know that you picked a queen on your first pick, that is a given and has noting to do with your future probabilities except decrease the number of queens and number of cards in the deck by 1.

Answer 10

so you can throw 4/52 out

Answer 11

now you have to pick the three next cards:
so, your probabilities will be, following the progression that you made:
3/51, 2/50, then 1/49

Answer 12

and you multiply them together to get your answer. Does that make sense?

Answer 13

i tried it that way but I still didn't get a correct answer

Answer 14

Hmm.. do you have the correct answer with you, by any chance?

Answer 15

no just the answer choices I can choose from but I think the answer is 1/425

Answer 16

Actually, looking back at the wording, the first pick counts as one of the three queens

Answer 17

So, you can throw out 1/49, which leaves you with 3/51*2/50

Answer 18

so it would be 3/51 times 2/51 which would give me 6/ 2550?

Answer 19

2/50, since you lose a card by picking a queen

Answer 20

but yeah 6/2550 sounds right

Answer 21

ok. thanks. do you think you can help me with another one. I haven't fully grasped the concept yet

Answer 22

Sure!

Answer 23

Two number cubes are rolled for two separate events:

All of the combinations of numbers on both cubes that give a sum less than 10.
All combinations of numbers on both cubes that give a sum that is a multiple of 3.

Answer 24

i have to Complete the conditional-probability formula for event B given that event A occurs first by writing A and B in the blanks

Answer 25

ok the formula is:

P(B|A)=P(A and B)/P(A)

Answer 26

and can be represented as an intersection, which is an upside down U

Answer 27

the word "and" to be clear

Answer 28

i'm not sure how to fill in the equation though

Answer 29

For this example?

Answer 30

yea with the number cubes

Answer 31

number cubes? as in die?

Answer 32

I'm guessing but the question said number cube

Answer 33

isnt number cube same as die?

Answer 34

I'm pretty sure

Answer 35

so lets say it has number 1-6

Answer 36

ok.. how would i go about figuring out part a to this All of the combinations of numbers on both cubes that give a sum less than 10.

Answer 37

find every combinations that is a multiple of 3 and is less than 10

Answer 38

for the multiple of 3 would it just be 2/6

Answer 39

(1,2), (2,1), (4,2),(2,4),(3,3),(5,1),(1,5)(5,4)(4,5)(6,3)(3,6)

Answer 40

11/36

Answer 41

I'm a lot lost when it comes to answering the question

Answer 42

multiple choice?

Answer 43

no its a fill in.

Answer 44

Could you provide the entire question?

Answer 45

Two number cubes are rolled for two separate events:

All of the combinations of numbers on both cubes that give a sum less than 10.
All combinations of numbers on both cubes that give a sum that is a multiple of 3.
Complete the conditional-probability formula for event B given that event A occurs first by writing A and B in the blanks:

Answer 46

So you basically have to consider the possibilities for each of these scenarios:

Lets consider A:
All of the combinations of numbers on both cubes that give a sum less than 10.

We know that there are many less combinations that sum to >=10 than those that are less, so we can look at the ones that are greater or equal to.
those combinations would be: (6,6), (6,6) (the other way), (6,5), (5,6), (6,4), (4,6), (5,5), (5,5)

So those are 8/36 total combinations that are >=10. To find those that are <10, we simply subtract this value from 1, so 1-8/36=28/36

Answer 47

We can use the formula to solve this, but it is much more effective to solve it logically

Answer 48

so how would I set it up for a conditional equation

Answer 49

well you have to find P(A andB) and P(A)

Answer 50

We found P(A), thats 28/36

Answer 51

P(A and B) is the probability of getting a sum that is a multiple of 3 and is less than 10.

Basically, the probability of getting a sum of 3, 6, or 9

Answer 52

9/36?

Answer 53

So you have to find the total number of combinations within those

a sum of 3 results from (1,2) or (2,1)
A sum of 6 results from (1,5), (5,1), (4,2), (2,4), (3,3), and (3,3)
9 results from (3,6), (6,3), (4,5), (5,4)

So I get 12/36 as P(A and B)

Answer 54

So my calculations would then lead to (12/36)/(28/36), which is 12/28, or 3/7

Answer 55

idk how to fit the answers in to the blanks though

Answer 56

the blanks are for P(A) and P(B)?

Answer 57

the blanks look weird

Answer 58

the blanks want me to fill in the equation

Answer 59

well the equation is P(B|A)=P(A and B)/P(A)
so, filled in, it would be:
P(B|A)=(12/36)/(28/36)

Answer 60

hey hey but isnt 3,3 only one

Answer 61

yes, but i would assume you count it twice since you can get it 2 ways, so it should be regarded as such when dealing with probability

Answer 62

oh ok

Answer 63

so we have to multiply 12/36 by 28/36?

Answer 64

You would have to divide 12/36 by 28/36

Answer 65

6/18*14/18= 3/9*7/9

Answer 66

oh ok

Answer 67

3/9*9/7

Answer 68

3/1*1/7

Answer 69

3/7?

Answer 70

Yes

Answer 71

wheres the owner of this forum