Hello, I need to learn how to solve problems similar to those I'm posting for my Probability and Statistics exam. Can you please explain to me which formulas I need to use to solve these problems?

Problem 1

Three six sided dice are thrown
a) What's the probability that 2 even and 1 uneven numbers come up
b) What's the probability that at least 2 even numbers come up
c) What's the probability that 2 sixes come up
d) What's the probability that more than or equal to 2 sixes come up

Problem 2

Five six sided dice are thrown
a) What's the probability that all numbers come up even
b) What's the probability that at least 1 six comes up
c) What's the probability that there is exactly 1 six
d) What's the probability that five different numbers come up

Problem 3

There are 2 urns with white and black balls. In the first urn there are 7 white and 8 black balls, in the second urn there are 9 white and 8 black balls.
From the first urn to the second 4 balls are moved. From the second urn 2 balls are taken out.
a) What's the probability that 2 white balls are taken out from the second urn
b) What's the probability, if 2 white balls are taken out, that 2 white and 2 black balls are moved

Problem 4

X; F(x) = {0, x <= 0
{1/δ'λ^7, x ∈ [0,2]
{1, x >= L
a) F(λ) = ?
b) D(x) = ?
c) p(x > L)

Forget about formulas. When you know how to think about these (at least 1-3), you won't need formulas.

Let's start with 1a: what's your thought process?

He just wants homework answers not tutoring.

I was thinking of answering these using Combinatorics but decided against it since I didn't want to be yelled at by David Sklansky for not doing it the easy way by using fractions.

Problem 1

Three six sided dice are thrown
a) What's the probability that 2 even and 1 uneven numbers come up
111
112
113
114
115
116

122 *
123
124 *
125
126 *

132
133
134
135
136

142 *
143
144 *
145
146 *

152
153
154
155
156

162 *
163
164 *
165
166 *

222
223 *
224
225 *
226

233
234 *
235
236 *

244
245 *
246

255
256 *

266

333
334
335
336

343
344 *
345
346 *

355
356

366 *

444
445 *
446

455
456 *

466

555
556

566 *

666

67 combinations/21 2 even 1 uneven combinations = 3,19047619047619

Yikes, we basically have to start from square one, but you've come to the right place, so panic not. For starters, you meant to say 21/67, because you can have a 31.34% chance but you can't have a 319% chance!

.3134 is still wrong, though, mainly because you tallied the combinations instead of permutations. (You also double-counted 11 of them, because the total should have been 56. For instance you counted 123 and 132 separately, but they're the same combination.)

Even though the order doesn't matter for this problem, you can't use combos to calculate a probability when repetitions are possible. If you do, the permutations involving repeats will be over-represented. For example, there are 6 arrangements of 123, but there is only one arrangement of 555. If you use combos, 123 gets the same treatment as 555, which is bad. The only way around that is to use permutations.

But there are 6^3 = 216 permutations and listing them all out would be a royal pain in the ass. Making matters worse, you probably wouldn't get any credit for it on an exam!

Instead, we can make life much easier by using some mathematical shortcuts.

First, take into account that we don't care about the actual numbers that come up, only their parity (even/oddness). Since any given die has a 50% chance of landing odd or even, this problem reduces to flipping 3 coins (or equivalently, flipping one coin 3 times) and asking the probability of 2 heads and 1 tail.

Each flip has 2 possibilities, and there are 3 flips, so the total # of permutations is 2⋅2⋅2 or 2³

Next: how many ways are there to arrange HHT (or EEO)? Well, the T can be in one of 3 different places, so 3.

So the answer is 3/8

For 1b we take the same 3 HHT perms and then add the HHH, bringing the tally up to 4 successful perms and making the answer 1/2.

There is another way of thinking about it that will be helpful for some of the other problems: we can treat each die as a separate event and then calculate the compound probability. With that in mind, take a stab at #1c (which doesn't translate to fair coinflips).


I was gonna joke that technically it's even easier (in the same way) to use brute force, but then OP beat me to it and attempted brute force!

Find the product of the probability of each individual outcome.

Problem 1a)
(3/6 evens) * (3/6 evens) * (3/6 odds) = 12.5%

Problem 2a)
(3/6 evens)^5 = 3.125%

It doesn't have to be EEO; it can also be EOE or OEE.

3b is the only somewhat interesting question.

Ship this man the pot!

Who needs free tutoring when you can just drop the course?