Question

The diagram below shows the contents of a jar from which you select marbles at random.


a. What is the probability of selecting a red marble, replacing it, and then selecting a blue marble? Show your work.
b. What is the probability of selecting a red marble, setting it aside, and then selecting a blue marble? Show your work.
c. Are the answers to parts (a) and (b) the same? Why or why not?

Answer 1

can you tell me the probability of selecting a red

Answer 2

how many reds are in the jar?

Answer 3

4

Answer 4

and how many marbles are in the jar all together

Answer 5

16

Answer 6

good, so the probability of first selecting a red is
P(r) = red marbles / all the marbles

Answer 7

which is 4/16 right?

Answer 8

yeah

Answer 9

now can you tell me the probability of selecting a blue ? (after putting the red back in)

Answer 10

4/7

Answer 11

um , try that bit again

Answer 12

P(b) = blue marbles / all the marbles

Answer 13

7/16

Answer 14

Good

Answer 15

now


The probability of selecting a red and then a blue is
P(r) x P(b)

Answer 16

\[P(\color{blue}b,\color{red}r)=P(\color{blue}b)\times P(\color{red}r)\\\qquad\qquad=\frac4{16}\times\frac7{16}\]

Answer 17

i might be easier to simply the first fraction before cross multiplying

Answer 18

do you know what i mean?

Answer 19

ya

Answer 20

\[\frac4{16}=\frac4{4\times4}=\frac{\cancel4}{\cancel4\times4}=\frac14\]

Answer 21

\[=\frac4{16}\times\frac7{16}\\=\frac1{4}\times\frac7{16}\\=\]

Answer 22

7/64?

Answer 23

right!\[\large\color{red}\checkmark\]

Answer 24

okay. thank you!!

Answer 25

so that is part a) done .

Answer 26

for part b)

\(P(\color{red}r)\) will be the same

but

\(P(\color{blue}b)=\frac{\text{blue marbles}}{ \text{all the marbles except the red one you took out } }\)

Answer 27

what do you get for P(b)?

Answer 28

7/16

Answer 29

not quite , remember one of the red ones isn't counted

Answer 30

7/9

Answer 31

7/15

Answer 32

thats it

Answer 33

right, so now P(b,r)= ?

Answer 34

7.4

Answer 35

\[P(\color{blue}b,\color{red}r)=P(\color{blue}b)\times P(\color{red}r)\\\qquad\qquad=\frac1{4}\times\frac7{15}\\\qquad\qquad=?\]

Answer 36

7/60

Answer 37

good,

Answer 38

What do you think about c) ?

Answer 39

no

Answer 40

they are different

Answer 41

which is more likely to happen ?