Question

probability help please

Answer 1

think about what it says (pick a card at random) RANDOM is the hint

Answer 2

so it would be independent

Answer 3

picking random cards would be independant

Answer 4

good job

Answer 5

thank you, do you know anything about this one?

Answer 6

just a sec

Answer 7

NO!!!!

Answer 8

1/55 is what i think

Answer 9

Events are independent if the outcome of one event does not influence the outcome of the other event. On the other hand, if the outcome of one event has an effect on the outcome of the other event, then they are dependent events.

Answer 10

whaaaa

Answer 11

is this one right?

Answer 12

absolutely

Answer 13

The first event is picking a card from the 11 cards.
You do not replace the card.
Now you only have 10 cards left.
The second event is picking the second card.
Now you are picking a card out of 10 cards left.
Surely the first event will influence the outcome of the second event.
Therefore, these are DEPENDENT events.

Answer 14

@mathstudent55 Are you talking about my second or third screenshot? :)

Answer 15

as i said before there are two I and 11 cards total

Answer 16

This is a great example of why the rules should be followed, and you should have only one question per post.
The first question was whether the events are dependent or independent.
All I have written about above is the first question.

Answer 17

I thought you could do more than one question per post i'm sorry.

Answer 18

I will close this question

Answer 19

okay so the second problem I'll just repost the link

http://assets.openstudy.com/updates/attachments/573f3bcce4b0e6895a864c28-posies9876-1463762120620-capture.png

Answer 20

I see that you didn't know about 1 question per post.
Now you understand why it's good rule.
It causes less confusion.

Answer 21

You got my explanation on the first problem, right?

Answer 22

Yes it's dependent right?

Answer 23

and yeah I'm sorry I didn't understand but I won't do the bunch of questions on one post thing again

Answer 24

Correct. The events of the first questions are dependent.
The reason is that once the first card is drawn, since the card is not replaced, that changes the probabilities of the picking of the second card. When the outcome of one event changes the outcome of the next event, the events are dependent.
If the card were replaced after the first drawing, then the events would be independent.

Answer 25

Ok, now let's look at the second question.

Answer 26

In the second problem, you are picking a card out of 11 cards.
There is only one A.
What is the probability of picking the A out of the 11 cards?

Answer 27

1/11

Answer 28

because one A and 11 cards.

Answer 29

yes

Answer 30

Correct.
Once the first card is picked, the card is put back.
Now you have again 11 cards.
Now you need to choose the B.
What is the probability of picking a B out of the 11 cards

Answer 31

same, 1/11

Answer 32

How many B's are there?

Answer 33

oh wait, sorry. 2/11

Answer 34

Correct.
Since these events, picking an A and picking a B with card replacement, are independent events, the probability of one event following the other event is the product of the individual probabilities.

Answer 35

Now multiply the two probabilities to find your answer.

Answer 36

A 2/121

Answer 37

Correct.

Answer 38

Ready for the third problem?

Answer 39

Yes. :)

Answer 40

For problem 3, you need to pick two I's without replacement.
The first drawing, you want an I out of 11 cards.
What is the probability of picking an I out of 11 cards?

Answer 41

2/11

Answer 42

Correct.
Now you picked one I.
You do not replace the I.
You have 10 cards left and only one I left.
What is the probability of picking an I now for the second drawing?

Answer 43

1/10

Answer 44

Correct.
Now multiply the two probabilities and reduce the fraction if necessary.

Answer 45

1 /110

Answer 46

Are you sure?

\(\dfrac{2}{11} \times \dfrac{1}{10} = ~ ?\)

Answer 47

1/55

Answer 48

\(\dfrac{2}{11} \times \dfrac{1}{10} = \dfrac{2}{110} = \dfrac{1}{55}\)

You are correct.

Answer 49

Thank you so much for all your help.

Answer 50

I have a couple more questions ( ones I'm not sure how to solve and ones I need my work checked on ) but I'll start a new thread.

Answer 51

Ok. I'll look for your posts.