Vtyjkyuk

Consider the following experiment: There is an urn with $100$ balls, which can be either black or white.
The number $B$ of black balls is a random variable, with the uniform distribution on $ 0 , 1 ,..., 100 $ , that
is,
$P(B = 0) = P(B = 1) = ··· = P(B = 100) = frac1101$.

a. If we pick one ball at random from the urn, what is the probability of it being black?

b. Suppose that we pick one ball, and it is white. Then, without replacing the first ball, a second one
is picked. What is the probability of this second ball being black?

c. Suppose now that we have picked $50$ balls in sequence, without replacement, and that all are white.
What is the probability that the 51st ball picked is black?

d. Finally, suppose that we have picked $50$ balls in sequence, with replacement, and that all are white.
What is the probability that the 51st ball picked is black?

a.)

Probability the first pick is black:
$$frac1101*frac0100+frac1101*frac1100+...+frac1101*frac100100=frac1100 * 101sum_n=0^100n=frac505010100=
frac12$$

b.)

So I know P(First ball picked is black) = $frac12$ so P(First ball picked is White) = $1-frac12=frac12$

P(Second ball picked is black) =$$frac1101*frac0100+frac1101*frac1100+...+frac1101*frac99100=frac1100 * 101sum_n=0^99n=frac495010100=frac99202$$
If this is right then is the probability $frac99202*frac12=frac99404$?

c.)

So I assume this works:
$$
P(B = 0) = P(B = 1) = ··· = P(B = 100) = frac1101$$$$
Rightarrow P(W = 0) = P(W = 1) = ··· = P(W = 100) = frac1101 $$
So P(Picking 50 white balls in sequence) =
$$prod_m=0^49Bigg(frac1100*101sum_n=1^100-mnBigg)$$
Then if thats right.. P(Picking 51st ball as black)=
$$frac1100 * 101sum_n=0^51n=frac132610100=frac6635050$$
So then this probability is:
$$frac6635050prod_m=0^49Bigg(frac1100*101sum_n=1^100-mnBigg)$$

d.)

Is this just $P(textWhite ball)^50 * P(textBlack ball)=(frac12)^51$

Your answer to (a) looks correct: I would state it as $$sumlimits_w=0^100 frac1101 times fracw100 =frac12.$$

Your answer to (b) does not. I would state the probability of a white ball followed (without replacement) by a black ball as $$sumlimits_w=0^100 frac1101 times fracw100 times frac100-w99 =frac16.$$ Then the conditional probability of a black ball given the preceding ball (without replacement) was white is this joint probability divided by the answer to (a), i.e. $$frac1/61/2=frac13.$$

My answer to (c) would take a similar approach. I would state the probability of fifty white balls without replacement as $$sumlimits_w=50^100 frac1101 times fracw! / (w-50)! 100!/50! =frac151,$$ while I would state the probability of fifty white balls without replacement followed by a black ball as $$sumlimits_w=50^100 frac1101 times fracw! / (w-50)! 100!/50! times frac100-w50=frac12652.$$ Then the conditional probability of a black ball given the preceding fifty balls (without replacement) were white is this joint probability divided by the earlier answer, i.e. $$frac1/26521/51=frac152.$$

And so my answer to (d) would also take a similar approach to (a) and (b), though not quite so pretty. I would state the probability of fifty white balls with replacement as $$sumlimits_w=0^100 frac1101 times left(fracw100right)^50 approx 0.02477513,$$ while I would state the probability of fifty white balls with replacement followed by a black ball as $$sumlimits_w=0^100 frac1101 times left(fracw100right)^50 times frac100-w100 approx 0.0003651899,.$$ Then the conditional probability of a black ball given the preceding fifty balls (with replacement) were white is this joint probability divided by the earlier answer, i.e. about $$frac0.00036518990.02477513approx 0.01474.$$