Question

PROBLEM 5. A box contains 10 tickets labeled 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Draw four tickets and find the probability that the largest number drawn is 8 if:

(a) the draws are made with replacement.

(b) the draws are made without replacement.

PROBLEM 6. Suppose a bakery mixes up a batch of cookie dough for 1,000 cookies. If there are raisins in the dough, it's reasonable to assume raisins will independently have a .001 chance of ending up in any particular cookie (assuming that raisins don't clump together), so that the number of raisins in a cookie behaves like a binomial (n= # raisins in whole batch, p= .001 ) random variable.

(a). How many raisins should be put into a batch of dough to make the chance of a cookie containing at least one raisin very close to 99% ? Calculate this using the binomial distribution.

(b). If there is a 99% probability that a cookie contains at least one raisin, what is the probability that a cookie contains exactly one raisin? Exactly 4 raisins? Calculate these using the binomial distribution, with the n you found in part (a).

(c). Now assume that the number of raisins in a cookie is Poisson, with mean mu. What is the right value of mu that makes the probability of at least one raisin in a cookie equal to 99% ?

(d). Assuming that the number of raisins in a cookie is Poisson with the mu that you found in part (c), what is the probability that a cookie contains exactly one raisin? Exactly 4 raisins?

REMARK There's something a bit subtle going on in this cookie story. For example, if you put 2,000 raisins into the dough and make 1,000 cookies, with raisins independently equally likely to end up in any of the 1,000 cookies, then of course the sum over all 1,000 cookies of the raisins-per-cookie numbers will add up to exactly 2,000. However, the fractions of cookies with exactly k raisins will, with very high probability, be close to Poisson (mu=2) probabilities, so those fractions would behave as though the numbers of raisins in cookies were independent Poisson (mu=2) random variables. If the numbers of raisins per cookie really were independent Poisson (mu=2) random variables, then the total number of raisins would be Poisson with mean 2,000 (which is approximately normal, with mean 2,000 and standard deviation SQRT(2,000) = 44.72). It turns out that the exact joint behavior of the rasins per cookie numbers with 2,000 raisins total is the same as the joint behavior with independent Poisson (mu=2) raisins per cookie, conditional on the total number of raisins being exactly 2,000.

Answer 1

Ans 5 a) Since one draw has to be 8 our focus is that the other three draws should be less than 8 so that the given condition could be satisfied and 8 has 4 places to be in from which it can select one in 4C1 ways that is 4 ways

And the rest three places can be filled in 7 ways that is from 1,2 . . . , 7 which can be filled in 7 * 7 * 7 so the total number of ways we can fill with replacement to get 8 as the highest number is 4 * 7 * 7 * 7 ways = 1372 ways.

But it is not necessary that it has to be just one 8 it can be more so if there are 2 8's .

So 2 8's can choose 2 places out of 4 places in 4C2 ways

And the rest two places can be filled in 7 ways that is from 1,2 . . . , 7 which can be filled in 7 * 7 so the total number of ways we can fill with replacement to get 8 as the highest number is 4C2 * 7 * 7 ways = 294

Similarly if there are 3 eights in 4C3 * 7 ways = 4 * 7 ways = 28 ways

And 4 8's in only one way .

And the total numbers of drawing 4 tickets is 10 * 10 * 10 * 10

So the required probability = ( 1372 + 294 + 28 +1) / 10 * 10 * 10 * 10 = 0.1695

Ans b If the draws are without replacement then it mean it is basically taking out 4 tickets at once and the total number of ways of doing that is 10C4.

Here since it is without replacement so it is only possible to have 8 and since 8 is the maximum number so one ticket has to be 8 and the rest has to be number among 1,2.... , 7 and we have to take out 3 tickets which can be done in 7C3 ways

.So the required probability is 7C3 / 10C4 = 1/6 = 0.1667

Ans 6 a) Let t be the number of raisins in the cookie

So basically we have P(t>=1) = 0.99 where t will follow binomial distribution.

which we can write as 1 - P(t=0) = 0.99 or P(t=0) = 0.01

Now as we know t follows binomial distirbution so P(t=0) = = 0.01  

since for binomial distribution P(t=r) =

or 0.001⁰ * 0.999ⁿ = 0.01

Taking log both sides we get n * log(0.999) = log(0.01)

Therefore n = log(0.01) / log(0.999) = 4602.867 = 4603 approximately

Ans b The required probability is P(t=1) and since t is a binomial random variable

P(t=1) = = 0.0461 approximately

Second problem is P(t = 4) = = 0.0462

Ans c Let t be the number of raisins in the cookie

So basically we have P(t>=1) = 0.99 where t will follow poisson distribution.

which we can write as 1 - P(t=0) = 0.99 or P(t=0) = 0.01

Now we now that P(t=r) = since t follows Poisson Distribution

So P(t=0) = = = 0.01

Therefore we get = - ln(0.01) = 4.605 approximately

Ans d) The required probabiltiy is P(t=1) = where = 4.605 as calculated in (c)

= = 0.04605

And for the second part we require to calculate P(t=4) = = 0.1874 approximately