Question

A box contains 5 red marbles, 3 blue marbles and 2 yellow marbles. A marble is chosen at random and replaced. This selection process is completed eight times. Find the probability that exactly 4 reds are selected

Answer 1

Could someone please explain it to me? @satellite73

Answer 2

what is the probability you get an red marble on any one try?

Answer 3

Every time the marble is chosen, the probability is the same because marbles are replaced.
Therefore use of the binomial distribution is indicated.
Have you done binomial distributions?

Answer 4

1/10

Answer 5

oh hell no

Answer 6

sorry 5/10 = 1/2

Answer 7

would n= 5 and k=4?

Answer 8

and p=1/2

Answer 9

ok better
so half the time you get it and half the time you don't get it
compute
\[\binom{8}{4}\left(\frac{1}{2}\right )^{8}\]

Answer 10

\(n=8, k=4, p=1-p=\frac{1}{2}\)

Answer 11

so there is no need for bernoulli distribution. what do they exactly mean by this sentence "A marlbe is chosen at random and replaced"?

Answer 12

these are bernoulli trials and you do use the binomial distribution

Answer 13

ok

Answer 14

but since both \(p\) and \(1-p\) are \(\frac{1}{2}\) you don't need to write something silly like
\[\binom{8}{4}(\frac{1}{2})^4\times (\frac{1}{2})^4\]

Answer 15

so when they say "this selection process is completed 8 times" it mean we pick from the box 8 marbles at a time?

Answer 16

im starting to get it now

Answer 17

Each pick of the marble is a Bernoulli trial. But when you pick 8 times and want to find 4 success (red), you need the Binomial distribution @satellite73 gave you.

Answer 18

what if it asked me about the probability of 1 red marble and 2 blue marbles?

Answer 19

You pick one marble at a time, out of 8 marbles all the time. That is why the probability remains 1/2 for all 8 trials (because of replacement).

Answer 20

of obtaining*

Answer 21

ill try do it on my on

Answer 22

own*

Answer 23

and correct me later if i am mistaken

Answer 24

That will be a little more complicated, because binomial distribution does not apply any more when your successful event include more than one colour.

Answer 25

ok could you please show me how it is done

Answer 26

Binomial distribution applies only to Bernoulli trials.

Answer 27

@satellite73 's expression breaks down as follows:
\( C^n_r\ p^r\ q^{n-r} \)
and applies to n Bernoulli trials with r successes (and n-r failures) with a probability of success equal to p (and probability of failure q=1-p).

Answer 28

so how is it done if they were to ask me to find the probability of mixed marlbes?

Answer 29

In the given problem,
n=8 (trials)
r=4 (successes)
n-r=4 (failures)
p= 1/2 (probability of success)
q= 1-1/2= 1/2 (probability of failure)
and
\( C^n_r =\frac{n!}{r!(n-r)!} = \frac{8!}{4!4!}\)
So the final probability comes down to:
\( P(4\ reds) = \frac{8!}{4!4!}(\frac{1}{2})^4(\frac{1}{2})^4 \)

Answer 30

thanks guys for your help and time. i really appreciate it

Answer 31

You're welcome! Hope you get the hang of the binomial distribution.

Answer 32

but what if i was asked to find the probability of mixed marlbes? for instance 1 red and 2 blue?

Answer 33

how can it be solved

Answer 34

@satellite73

Answer 35

would it look like this (10 5) (10 3)/ (10 3)????

Answer 36

sorry (10 1) (10 2)/ (10 3)????

Answer 37

im stuffing up real bad :(

Answer 38

I do not know of an explicit formula for that, see if @satellite73 has a clue.
This problem will have to resort to counting, or probability trees. Have you done any of these?

Answer 39

i guess it is (5 1) (3 2) /(10 3)

Answer 40

yes with @satellite73

Answer 41

it should be 0.125

Answer 42

In this case, we can still use the binomial distribution to pick 5 yellows, namely
\( C^8_5\ (\frac{1}{5})^5 \ (\frac{4}{5})^3 \)
There are 8 possible arrangements for the remaining marbles, red and blue, namely
RRR
RRB
RBR
RBB
BRR
BRB
BBR
BBB
out of which 3 are 1 red and 2 blues.
So multiply the above by \( \frac{3}{8} \) to get
\( \frac{3}{8}\ C^8_5\ (\frac{1}{5})^5 \ (\frac{4}{5})^3 \)

Answer 43

one red and two blue out of how many picked? three?

Answer 44

yes

Answer 45

out 0f 10 marlbes in total

Answer 46

you had it above

Answer 47

that is how you taught me to do it

Answer 48

is it right?

Answer 49

\[\frac{\binom{5}{1}\binom{3}{2}}{\binom{10}{3}}\]

Answer 50

:)

Answer 51

thanks little by little im getting good at this :)

Answer 52

yeah practice is good

Answer 53

@satellite73 just to make sure, the solution is for picking 3 marbles, right?

Answer 54

yes

Answer 55

thank you!

Answer 56

yw
clear right? it is instant to write the solution, just takes a minute to compute

Answer 57

i got an ice chest with 7 miller, 8 bud, 10 strohs
if i pick 4 at random, the probability i get 1 miller 2 buds and a stohs is
\[\frac{\binom{7}{1}\binom{8}{2}\binom{10}{1}}{\binom{15}{4}}\]

Answer 58

Do you mean C(25,4) at the bottom?