Question

Sam picked a card from a standard deck, looked at it, and then put it back. He then picked a second card. What is the probability that Sam picked a heart or a king on either pick?

one over thirteen

sixteen over fifty two

seventeen over fifty two

sixteen over fifty three

Answer 1

@Kainui

Answer 2

@satellite73

Answer 3

@zepdrix

Answer 4

How many hearts and how many kings in a standard deck of cards?

Answer 5

4 kings and 15 hearts

Answer 6

*13 hearts

Answer 7

So total how many hearts and kings?

Answer 8

17

Answer 9

which is basically 17/52

Answer 10

Think about 17 is that correct?

Answer 11

I am not sure.

Answer 12

i think there are 13 hearts and and another 3 kings for a total of \(16\) not \(17\) to pick from

Answer 13

OK there are 13 hearts (A,2,3,4,5,6,7,8,9,10,JQ,K) that is correct and there are 4 kings (KS,KC,KD,KH) and that is correct....but have we counted the king of hearts twice?

Answer 14

Yeah

Answer 15

then it is a matter of how to compute "on either pick" meaning on pick one, pick two, or both

Answer 16

for this i would use
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]

Answer 17

Would you mind helping me for this one?

Answer 18

actually i don't see the answer i would get as one of the choices
maybe i am doing it wrong

Answer 19

You sure?

Answer 20

probability he gets it on the first pick is \(\frac{4}{13}\) right?

Answer 21

yes

Answer 22

and since he replaces the card, the probability he gets it on the second pick is also \(\frac{4}{13}\)

Answer 23

Yeah I believe so.

Answer 24

now you are asked the probability he gets it on either pick
in math that should mean on the first pick, on the second pick or on both picks

Answer 25

no the probability is 16/52

Answer 26

13 hearts and 3 kings

Answer 27

for that we would use
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\] with
\[P(A)=P(B)=\frac{4}{13}\] and
\[P(A\cap B)=(\frac{4}{13})^2\]

Answer 28

the last one because they are independent
@nelsonjedi \(\frac{16}{52}=\frac{4}{13}\)

Answer 29

My bad

Answer 30

It is 16/53.

Answer 31

ooh i see!! what a poorly worded question!

Answer 32

when you are asked "on either pick" they do not mean when you pick twice!! they just mean what is it on a pick!

Answer 33

Oh nevermind sorry I was looking at another response

Answer 34

ok so it is \(\frac{4}{13}\) but the question was written strangely

Answer 35

the real answer is
\(\frac{88}{169}\) but they want \(\frac{16}{52}\) so give them that one

Answer 36

A poll was conducted in states A, B, and C to determine which candidate voters would most likely vote for. The results are displayed in the table below:


State A State B State C Total
Liberal 20 21 12 53
Conservative 20 14 13 47
Total 40 35 25 100


What is P(liberal | State B)?
0.21

0.35

0.4

0.6

Answer 37

@satellite73

Answer 38

Kris wanted to understand whether students at her school were in favor of an extended school day. She surveyed some students and displayed the results in the table below:


In favor Opposed Undecided
Grade 9 6 4 8
Grade 10 10 11 9
Grade 11 12 15 11
Grade 12 15 6 14


If the principal randomly selects a student in grade 10 from this survey, what is the probability that the student is opposed to extending the school day?
0.09

0.29

0.30

0.31

Answer 39

Do you think you can help @kropot72

Answer 40

@ganeshie8

Answer 41

@gabylovesu Are you sure that you have copied the table and the choices of answer correctly? The reason that I ask is that none of the choices appear to be correct.

Answer 42

Yup I did

Answer 43

The way I look at it is this. We are given that the randomly selected student is in grade 10. The required probability is therefore:
\[\frac{grade\ 10\ number\ against}{total\ grade\ 10\ students}\]

Answer 44

The function f(x) = 15(2)x represents the growth of a frog population every year in a remote swamp. Elizabeth wants to manipulate the formula to an equivalent form that calculates every half-year, not every year. Which function is correct for Elizabeth's purposes?

f(x) = 15(2 to the one half power)2x

f(x) = 15(22)the x over 2 power

f(x) = fifteen halves(2)x

f(x) = 30(2)x @ganeshie8

Answer 45

Izzy's heartbeat rate was measured on an EKG machine. After conversion, the function produced could be modeled by a sine function, and the wave produced a maximum of 8, minimum of −2, and period of ð. Which of the following functions could represent Izzy's EKG read-out?

f(x) = 8 sin ðx − 2

f(x) = 5 sin ðx + 3

f(x) = 5 sin 2x + 3

f(x) = 2 sin 2x + ð

Answer 46

@ganeshie8

Answer 47

@kropot72 do you think if the student was randomly selected then should consider the ways of choosing the opposed 10th grade student from the total number surveyed

I think this has combined probability
grade and how voted

41 opposed/total # surveyed and 11/30 10th grade opposed

Answer 48

@gabylovesu Kris wanted to understand whether students at her school were in favor of an extended school day. She surveyed some students and displayed the results in the table below:
I think the answer should be 0.09

11- 10th grade opposed from 121 surveyed
11/121 = 0.09

Answer 49

@triciaal
It is known that the randomly selected student is in grade 10. What is the probability that a student in grade 10 is opposed? Surely the answer is the fraction (total number opposed in grade 10) divided by (total number in grade 10):
\[\frac{11}{30}=0.367\]
Lets look at the probability that a randomly selected student if opposed:
\[P(opposed)=\frac{total\ opposed}{total\ students}=\frac{36}{121}=0.2975\]
On the basis of the above, do you really think that 0.09 can be correct?

Answer 50

yes because there are still 11 random in grade 10

Answer 51

looking at 2 conditions grade 10 and opposed

Answer 52

My understanding of the question is that the principal goes to grade 10 and randomly selects a student. Therefore only the students in grade 10 are to be considered in the calculation.

Answer 53

the principal randomly selects a student in grade 10

grade 10 and opposed 30/121*11/36 = 0.07 very close to 0.09 rounding maybe

Answer 54

English is difficult the agreed solution.

Answer 55

It is not possible for there to be only a 9 in 100 chance for the student to be opposed.

Answer 56

not just opposed!!! 2 conditions are met #1 10th grade and #2 opposed

Answer 57

But the question already gives the information that the student is in grade 10. Therefore the probability that the principal selects a student who is in grade 10 does not need to be calculated.

Answer 58

yes because he was randomly selected from the different grades that were in the survey the first condition. I am serious when I said English is difficult. Read again how it is written. I understand your position.

Answer 59

by the way I wrote "he" that should be "the student" because it did not say if it was a boy or girl