# 3 Probability problems and 1 Statistics problem

#### whois1230

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)

#### romsek

Math Team
I think it's too likely that these are the actual problems from your test to just give you answers.

Sketch out some approaches to each problem and point out where you are stuck on each.

If you can't do at least this you probably deserve to fail your exam.

whois1230

#### whois1230

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

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

b) What's the probability that at least 2 even numbers come up

67/29 = 2,310344827586207

c) What's the probability that 2 sixes come up

67/5 = 13,4

d) What's the probability that more than or equal to 2 sixes come up

67/6 = 11,16666666666667

#### romsek

Math Team
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
Well... there are easier ways to go about it. And some of your answers show a definite lack of understanding.
Probabilities are always in [0,1]

a) $P[even]=P[odd]=1/2$

$P[\text{2 even, 1 odd}] = \dbinom{3}{2}(1/2)^2(1/2)^1 = \dfrac{3}{8}$

b) $P[\text{at least 2 even}] = P[\text{2 even}]+P[\text{3 even}] = \dbinom{3}{2}(1/2)^3 + \dbinom{3}{3} (1/2)^3 = \dfrac 4 8 = \dfrac 1 2$

c)
For a single roll $P[6] = 1/6$
$P[\text{2 6's out of 3 rolls}]=\dbinom{3}{2}(1/6)^2(5/6)^1 = \dfrac{15}{216} = \dfrac{5}{72}$

d)
$P[\text{2 or more 6's rolled in 3 rolls}] = P[\text{2 6's}]+P[\text{3 6's}] = \dfrac{5}{72} + \dbinom{3}{3}(1/6)^3 = \dfrac{16}{216} = \dfrac{2}{27}$

whois1230

#### romsek

Math Team
#2 is just like #1 so I'm going to let you do that one.

#3 is just a pain basically you start with (9w,8b) in the 2nd urn.
You move 4 balls from urn 1 to urn 2. These can be (0w,4b), (1w,3b), (2w,2b), (3w,1b), (4w,0b)

which result in urn 2 having (9,12), (10, 11), (11, 10), (12, 9), or (13,8) white and black balls.

$P[\text{we transfer m white balls and 4-m black balls to urn 2}] = \dfrac{\dbinom{7}{m}\dbinom{8}{4-m}}{\dbinom{15}{4}}$

$P[\text{we pick 2 white balls from urn 2 given we've transferred m white balls and 4-m black balls}] = \dfrac{\dbinom{9+m}{2}}{\dbinom{21}{2}}$

So combining all this

$P[\text{we pick 2 white balls from urn 2}] = \sum \limits_{m=0}^4 \dfrac{\dbinom{9+m}{2}}{\dbinom{21}{2}} \cdot \dfrac{\dbinom{7}{m}\dbinom{8}{4-m}}{\dbinom{15}{4}} = \dfrac{9}{35}$

whois1230

#### romsek

Math Team
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)
I can't understand this one. what is $D(x)$ ?

#### whois1230

Well... there are easier ways to go about it. And some of your answers show a definite lack of understanding.
Probabilities are always in [0,1]

a) $P[even]=P[odd]=1/2$

$P[\text{2 even, 1 odd}] = \dbinom{3}{2}(1/2)^2(1/2)^1 = \dfrac{3}{8}$

b) $P[\text{at least 2 even}] = P[\text{2 even}]+P[\text{3 even}] = \dbinom{3}{2}(1/2)^3 + \dbinom{3}{3} (1/2)^3 = \dfrac 4 8 = \dfrac 1 2$

c)
For a single roll $P[6] = 1/6$
$P[\text{2 6's out of 3 rolls}]=\dbinom{3}{2}(1/6)^2(5/6)^1 = \dfrac{15}{216} = \dfrac{5}{72}$

d)
$P[\text{2 or more 6's rolled in 3 rolls}] = P[\text{2 6's}]+P[\text{3 6's}] = \dfrac{5}{72} + \dbinom{3}{3}(1/6)^3 = \dfrac{16}{216} = \dfrac{2}{27}$
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

a) (3/6)^5 = 3.125%
b) (1/6)^5 = 0.00012860082 = 0.01286008%
c) (N/K)(1/6)^K(5/6)^N-K
(5/1)(1/6)^1(5/6)^5-1 = 5(1/6)(5/6)^4 = 5/6(5/6)^4
d) 5/6*4/6*3/6*2/6*1/6 = 120/7776 = 0.015
5!/6^5 = 120/7776 = 0.015

Can you please tell me which formula you used for Problem 1 c and d?

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#### romsek

Math Team
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

a) (3/6)^5 = 3.125%
b) (1/6)^5 = 0.00012860082 = 0.01286008%
c) (N/K)(1/6)^K(5/6)^N-K
(5/1)(1/6)^1(5/6)^5-1 = 5(1/6)(5/6)^4 = 5/6(5/6)^4
d) 5/6*4/6*3/6*2/6*1/6 = 120/7776 = 0.015
5!/6^5 = 120/7776 = 0.015

Can you please tell me which formula you used for Problem 1 c and d?
b) is incorrect. The formula you used corresponds to the probability that 5 sixes were rolled.

Hint: $P[\text{at least 1 six was rolled}] = 1 - P[\text{0 sixes were rolled}]$

c) you look like you're just parroting a formula w/o understanding it. Here $N=5,~K=1,~p=1/6$ and you use the binomial distribution formula

$$\displaystyle p(K) = \dbinom{N}{K}p^K (1-p)^{N-K}$$

d) is almost correct. The answer is $1 \cdot 5/6 \cdot 4/6 \cdot 3/6\cdot 2/6 = \dfrac{120}{6^4} = \dfrac{5}{54}$

Answering your question about 1(c), (d), I gave you the formula for $p(k)$ for a given $k$ above.
If the question asks for a range, or variety of $k$'s you just have to sum up the probabilities of each $k$ that is included.

whois1230

#### whois1230

b) is incorrect. The formula you used corresponds to the probability that 5 sixes were rolled.

Hint: $P[\text{at least 1 six was rolled}] = 1 - P[\text{0 sixes were rolled}]$

c) you look like you're just parroting a formula w/o understanding it. Here $N=5,~K=1,~p=1/6$ and you use the binomial distribution formula

$$\displaystyle p(K) = \dbinom{N}{K}p^K (1-p)^{N-K}$$

d) is almost correct. The answer is $1 \cdot 5/6 \cdot 4/6 \cdot 3/6\cdot 2/6 = \dfrac{120}{6^4} = \dfrac{5}{54}$

Answering your question about 1(c), (d), I gave you the formula for $p(k)$ for a given $k$ above.
If the question asks for a range, or variety of $k$'s you just have to sum up the probabilities of each $k$ that is included.
b) (5/6)^5 = 5^5/6^5 = 3125/7776
1 - 3125/7776 = 4651/7776 ~ 0.598
c) p(K) = (N/K)p^K(1-p)^N-K
(5/1)1/6^1(1-1/6)^5-1
5/6(5/6)^4

Sorry, I didn't understand how you used the formula for d)

#### romsek

Math Team
Sorry, I didn't understand how you used the formula for d)
There's not much "formula" to it.
Imagine you throw the dice one at a time.
The first number can be anything, i.e. probability 1
The second number must be different from the first, i.e. probability 5/6
The third different than both of the first two, i.e. probability 4/6,
etc. continuing until there are only 2 choices for the last die roll.

Combine these you get $P[\text{5 different #s}] = 1 \cdot \dfrac{5}{6}\cdot \dfrac{4}{6}\cdot \dfrac{3}{6}\cdot \dfrac{2}{6} = \dfrac{5}{54}$

whois1230