Question

PLEASE HELLLPPP!! I will give medal!
Two cards are drawn from a standard deck of cards.

Part A: If they are drawn with replacement, what is the probability that both cards are 2s?

Part B: If they are drawn without replacement, what is the probability that the first card is a club and the second card is a spade?

Part C: Which of the two scenarios in Part A or Part B represents dependent events? Explain.

Answer 1

What about it do you not understand?

Answer 2

I kind of need everything explained. Maybe go step by step @legendarysadist

Answer 3

Very few people on here will go through a step-by-step explanation of everything. A base knowledge of the subject is first required before you ask a certain question. If it is probability you are confused with, go review it before asking questions. If it is the wording of the question, what about it confuses you?

Answer 4

The probability is what confuses me, how exactly do I find that?

Answer 5

Probability is amount you want/amount total. If there are 10 cards, and 5 are red, 5 are blue, what is your chance of getting a red card?

Answer 6

50%?

Answer 7

Yes. So if there are 13 cards, and 1 of them is a 2, what is your chance of getting a 2?

Answer 8

6%?

Answer 9

You can just keep it in fraction form, so 1/13 (and it would be 8% anyways)

Answer 10

Now let's say you return the 2 back to the deck, what is your chance of getting it again?

Answer 11

Ok so if you have 2 cards out of 52, with replacement, does that mean you divide 2/52?

Answer 12

If you are taking a card twice, that means you multiply the probabilities of getting it both times.

Answer 13

I don't want to get you confused, so let's start at the beginning

Answer 14

In a standard deck, there are 4 2's and 52 cards total. So what would that ratio be simplified?

Answer 15

It would be 2?

Answer 16

4/52 simplifies to what?

Answer 17

7% or 0.07

Answer 18

AGAIN KEEP IT AS A FRACTION. WE ARE NOT DIVIDING.

Answer 19

Okay sorry 1/13

Answer 20

As a reminder, we are going to stay in fractions always, unless the question asks for the answer as a decimal or percentage. Fraction should be the default. But let's continue

Answer 21

okay got it :)

Answer 22

So the chance of getting a 2 is 1/13. Now if we put the card back into the deck, will we have the same chance of getting the 2 or a different one?

Answer 23

We will have the same chance.

Answer 24

Good. So if we want to get the chance of having both happen, what we do is multiply the probability. So 1/13 * 1/13 = ?

Answer 25

so 1/169

Answer 26

So this is if they are drawn with replacement?

Answer 27

Correct. That is our answer for the first problem. Yes, that is right.

Answer 28

Okay!

Answer 29

So what differences do you think there will be in solving it when we do the second problem?

Answer 30

Since it's without replacement, there will be a change in the probability

Answer 31

Correct. Anything else?

Answer 32

I am not sure, this is the one that really confused me.

Answer 33

Well do you know how many spades there are in a deck? What about clubs?

Answer 34

there are 13 I think

Answer 35

That is correct. So what would the probability be of getting a spade out of all of the cards in the deck?

Answer 36

so would we do 13/52?

Answer 37

That would find us the probability of getting a spade, yes.

Answer 38

1/4 would be the probability

Answer 39

Correct. Now we took 1 card out of the deck, so how many cards in the deck are there now? (Since we aren't replacing the card)

Answer 40

13/51

Answer 41

That doesn't simplify any further either.

Answer 42

Good. So now we will multiply the probabilities together.

Answer 43

13/204 would be the final answer?

Answer 44

Yes, that would be it

Answer 45

Oh wow that made it so much easier. I have a few more, do you have time to help?

Answer 46

Sure, I can do a couple more

Answer 47

A manufacturer produces soda cans and a quality control worker randomly selects two cans from the assembly line for testing. Past statistics show that 10% of the cans are defective. What is the probability that the two selected cans are defective if the quality control worker selects the two cans from a batch of 60 cans?

P(Both defective) = 6/25
P(Both defective) = 1/118
P(Both defective) = 3/250
P(Both defective) = 9/625

Answer 48

Any ideas on how you would do this?

Answer 49

We can start with 2/60?

Answer 50

10% of all the cans are defective. So for 60 cans, how many are defective?

Answer 51

6 cans are defective

Answer 52

So for the first can, what is the chance of it being a defective can?

Answer 53

6/60 = 1/10

Answer 54

Good. Now for the second can, we are now missing 1 can out of the defective cans, and one can out of the total cans. So what would the new probability be for the second can?

Answer 55

5/59? I am not sure about that one

Answer 56

Then we multiply

Answer 57

Well you got it, so no worries. Now multiply the two probabilities together to get your answer :)

Answer 58

1/118 is the answer! Thanks.
Here is the next one!
Consider two events such that P(A) = 1/4, P(B) = 2/3, and P(A ∩ B) = 1/5. Are events A and B independent events?

Yes, they are independent because P(A) ⋅ P(B) = P(A ∩ B)
No, they are dependent because P(A) ⋅ P(B) = P(A ∩ B)
Yes, they are independent because P(A) ⋅ P(B) ≠ P(A ∩ B)
No, they are dependent because P(A) ⋅ P(B) ≠ P(A ∩ B)

Answer 59

Any ideas?

Answer 60

Okay so if they are independent, that means that they are with replacement, right?

Answer 61

Yes, that is pretty much what it means.

Answer 62

Well the first two choices are out because 1/4 * 2/3 does not equal 1/5

Answer 63

@legendarysadist

Answer 64

Yes, you are on the right track

Answer 65

So how do I know if it's dependent or independent?

Answer 66

Well I could explain this for a while but simply, if you cannot multiply them together then that means something must be changed in them, so they must be dependent of each other.

Answer 67

Oh I got it! You have no idea how much you are helping me! Thank you thank you thank you

So I just have one more! You still have time?

Answer 68

Yeah I can do one more

Answer 69

At an amusement park, the probability that a child eats a hot dog and drinks a soda pop is 0.48. The probability that a child eats a hot dog is 0.59, and the probability that a child drinks soda pop is 0.88. What is the probability (rounded to the nearest hundredth) that a child drinks soda pop given that the child has already eaten a hot dog?


0.55
0.81
0.28
0.42

Answer 70

Just multiply the probabilities together.

Answer 71

I think it's the third choice. 0.28

Answer 72

oh wait

Answer 73

I think I misread

Answer 74

OK

Answer 75

Ok, so the child eating a hot dog is for certain, so we will not deal with it. The two variables we have left are that he will drink soda, and the chance of him doing both. So we will multiply them together.

Answer 76

0.88 * 0.59?

Answer 77

.88 * .48

Answer 78

No it's 0.48*0.88

Answer 79

and the answer is 0.42

Answer 80

Yes, that is correct. The way they ask this one is a bit confusing tbh

Answer 81

Yeah tell me about it! Thanks for all your help!!!!!

Answer 82

Haha no problem (: