Question

A bag contains 4 white and 6 black marbles. 28 students each reach in, select a marble, record its color, then replace the marble.
a. What is the probability that exactly 10 students select white marbles?
b. What is the probability that no more than 10 students select white marbles?

Answer 1

binomial distribution

Answer 2

?

Answer 3

\[P(10/white)=\frac{28!}{10!18!}(0.4)^{10}(0.6)^{18}\]

Answer 4

they want "no more than 10" so they will need to do a little more work.

Answer 5

true, been stuck on this for ages...

Answer 6

Let P(10 white) be the probability that 10 students pick white.
Probability that no more than 10 pick white is [1 - P(10 white)].
@uj1024 What do you get for P(10 white) ?

Answer 7

that is not quite right

Answer 8

9/14?

Answer 9

I get 0.14 for the probability that 10 will pick white using the equation that I posted earlier.

Answer 10

UPDATE!!!
i added question a. as a hint, although it has already been solved

Answer 11

\[\sum_{x=0}^{10}{28\choose x}(0.4)^{x}(0.6)^{28-x}\]

Answer 12

i dont understand the difference between exactly and no more than

Answer 13

i dont think sequences and sigma notation have anything to do with this...

Answer 14

no more than 10...means 10 is the max...so you can have anywhere between 0 and 10 inclusive

Answer 15

so, 10! ?

Answer 16

I suggest you read up on the binomial distribution

Answer 17

is that related to binomial theorem? i havent i learnt about binomial distribution

Answer 18

yes

Answer 19

@uj1024 What Zarkon has said is right. Please read up on the binomial distribution and then you should be able to understand the solution that he has given.

Answer 20

For each student, let the probability of picking a white marble be p, and the probablity of picking a black marble be q. (Note that q = 1 - p by definition.) The probability that the sequence picked will be

wwbwwbwwbwwbwwbwwbwwbwwbwwb

is
\[p^{19} q^9\]
Because white was drawn 19 times and black 9 times. Of course, this also the probality of drawing bwwbwwbwwbwwbwwbwwbwwbwwbww, or indeed any sequence that involves 19 whites and 9 blacks. How many such sequences are there? The number of ways of arranging 28 objects, 19 of which are identical (white) and another 9 of which are identical (black), or 28!/(19! 9!). The probability of getting any one of these sequences is the sum of all terms, or:

\[\frac{28!}{19! 9!} p^{19} q^9\]

We can immediately generalize to say the probability of getting any of the possible sequences of k whites and n-k blacks is:
\[\frac{ n! }{k! (n-k)!} p^k q^{n-k}\]
What you need is the probability that at least 18 students pick black marbles, so you need the sum of this expression from k = 0 (no white marbles) to k = n - 18 (all but 18 marbles white):
\[\sum_{k=0}^{n-18} \frac{n!}{k! (n-k)!} p^k q^{n-k}\]

Unfortunately, I know of no closed expression for this object. If I write a program to calculate this sum, for p = 4/10, n = 28, using the GNU GMP program, I get:
\[\frac{1486137203415528633}{1490116119384765625}\approx 0.997330\]

Answer 21

Oops, I typed in the wrong arguments (28 and 18) to my program. For n = 28 and kmax = 10 I get:
\[\frac{2969550422392190877}{7450580596923828125}\approx 0.398566\]
which seems just a bit more reasonable. Also, this is the GNU GMP library.

Answer 22

as I wrote above


\[\sum_{x=0}^{10}{28\choose x}(0.4)^{x}(0.6)^{28-x}\]

which equals \[\frac{2969550422392190877}{7450580596923828125}\]

:)

Answer 23

Yes, I agree. I gave the longer explanation so that the OP could see WHY this is the correct solution. How did you get the numerical value, by the way?

Answer 24

I just used my calculator

Answer 25

Goodness, they make remarkable calculators these days. The last one I used was the venerable HP-41.

Answer 26

I have a TI-Nspire CX CAS

Answer 27

http://en.wikipedia.org/wiki/File:TI-nspire_CX_CAS.jpg

Answer 28

Interesting, thanks. I haven't owned a calculator in probably 15 years, since I'm usually never far from a computer, and it takes very little time to write a little C code to do anything more complex than 1 + 1. But the ability to easily do arithmetic with more than 15 digits precision is intriguing..

Answer 29

whenever I need to program something I use MatLab