The University of California, Berkeley, stat2.2x probability the probability of a preliminary study note: midterm

The stat2.2x probability (probability) course was taught at the EdX platform in 2014 by the University of California, Berkeley (University of California, Berkeley).

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Practice problems for the midterm

Problem 1

In a group of 5 of the high school students, 2 am in 9th grade, 2 am in 10th grade, and 1 was in 12th grade. The students is picked at random without replacement.

A) The first student picked is not in 10th grade

b) Both of the students is in 9th grade

c) The second student picked is in 12th grade

D) The second student picked is in 12th grade, given the first student picked are in 9th grade

e) At least one of the students are in 10th grade

f) The students is at least, grades apart

Solution:

A) $ $P (\text{first student is not in 10th grade}) =\frac{3}{5}$$

b) $ $P (\text{both is in 9th grade}) =\frac{2}{5}\times\frac{1}{4}=\frac{1}{10}$$

c) $ $P (\text{second student is in 12th grade}) =\frac{4}{5}\times\frac{1}{4}=\frac{1}{5}$$

d) $ $P (\text{second student is in 12th grade | First Student are in 9th grade}) $$ $$=\frac{p (\text{first are in 9th grade &A mp Second is in 12th grade})}{p (\text{first student are in 9th grade})}=\frac{\frac{2}{5}\times\frac{1}{4}}{\frac{2}{5}}=\ frac{1}{4}$$

E) $ $P (\text{at least one student is in 10th grade}) $$ $$=1-p (\text{no student are in 10th grade}) =1-\frac{3}{5}\times\frac {2} {4}=\frac{7}{10}$$

f) $ $P (\text{at least, Grades apart}) =p (\text{grade 9 &) +p (\text{grade &) $$ $$= (\frac{2}{5}\times\fr AC{1}{4}+\FRAC{1}{5}\TIMES\FRAC{2}{4}) + (\frac{2}{5}\times\frac{1}{4}+\frac{1}{5}\times\frac{2}{4}) =\frac{2}{5} $$

Problem 2

In a raffle there was tickets of which 1 is the winner. I pick 2 tickets at random. Find the chance in least one of the tickets that I pick is the winner, if I pick

A) with replacement

b) without replacement

Solution:

A) $ $P (\text{at least one is the winner with replacement}) $$ $$=1-p (\text{none of the The and the winner}) =1-\frac{99}{100} \times\frac{99}{100}=0.0199$$

b) $ $P (\text{at least one is the winner without replacement}) $$ $$=1-p (\text{none of the And the winner}) =1-\frac{99}{1 00}\times\frac{98}{99}=0.02$$

Problem 3

In an undergraduate class, 70% of the students is juniors and the rest are seniors. Of the juniors, 40% is math majors. Of the seniors, 25% is math majors. One student is picked at random. Find

A) $P (\text{junior}) $

b) $P (\text{senior}) $

c) $P (\text{math major | junior}) $

d) $P (\text{not a Math major | junior}) $

e) $P (\text{math major | senior}) $

f) $P (\text{not a Math major | senior}) $

g) $P (\text{junior Math major}) $

h) $P (\text{senior Math major}) $

i) $P (\text{math major}) $

j) $P (\text{junior | Math major}) $

k) $P (\text{senior | Math major}) $

L) $P (\text{junior | Not a Math major}) $

Solution:

A) $ $P (\text{junior}) =0.7$$

b) $ $P (\text{senior}) =1-0.7=0.3$$

c) $ $P (\text{math major | junior}) =0.4$$

d) $ $P (\text{not a Math major | junior}) =1-0.4=0.6$$

E) $ $P (\text{math major | senior}) =0.25$$

f) $ $P (\text{not a Math major | senior}) =1-0.25=0.75$$

g) $ $P (\text{junior Math major}) =0.7\times0.4=0.28$$

h) $ $P (\text{senior Math major}) =0.3\times0.25=0.075$$

i) $ $P (\text{math major}) =p (\text{junior Math Major}) +p (\text{senior Math major}) $$ $$=0.28+0.075=0.355$$

J) $ $P (\text{junior | Math major}) =\frac{p (\text{junior Math Major})}{p (\text{math major})}=\frac{0.28}{0.355}\ doteq0.7887324$$

k) $ $P (\text{senior | Math major}) =\frac{p (\text{senior Math Major})}{p (\text{math major})}=\frac{0.075}{0.355}\ doteq0.2112676$$

L) $ $P (\text{junior | Not a Math major}) =\frac{p (\text{junior not a math major})}{p (\text{not a math major})}$$ $$=\frac{0 .7\times (1-0.4)}{1-0.355}\doteq0.6511628$$

Problem 4

There is both events, $A $ and $B $. $P (A) =0.2$ and $P (B) =0.3$.

A) are $A $ and $B $ independent?

i) Yes. II) No. iii) Maybe, or maybe not; There is not enough information to decide.

b) What's the biggest that $P (\text{a or B}) $ can be?

c) What's the smallest that $P (\text{a or B}) $ can be?

Solution:

A) Maybe, or maybe not; There is not enough information to decide.

b) $ $P (\text{a or B}) =p (a) +p (b)-P (a \cap B) \le P (a) +p (b) =0.5$$

c) $ $P (\text{a or B}) \ge \max\{p (A), P (b) \}=0.3$$

Problem 5

There is both events, $A $ and $B $. $P (A) =0.2$, $P (b) =0.3$, and $P (\text{a or B}) =0.4$. are $A $ and $B $ independent?

i) Yes. II) No. iii) Maybe, or maybe not; There is not enough information to decide.

Solution:

No. $ $P (a \cup B) =p (a) +p (b)-P (a\cap b) =0.2+0.3-p (A\cap b) =0.4$$ $$\rightarrow p (a\cap b) =0.1\neq P (a) \cdot p (b) =0.06$$

Problem 6

A die is rolled repeatedly. Find the chance that

A) Each of the first three rolls shows one spot

b) The first three rolls all show the same face

c) The face and six spots appears at most once among the first 5 rolls

D) The face and six spots appears for the first time in the 5th roll

e) The face and six spots appears for the second time in the 5th roll

Solution:

A) $ $P (\text{each of the first three rolls is one spot}) = (\frac{1}{6}) ^3=\frac{1}{216}$$

b) $ $P (\text{each of the first three rolls is the same spot}) =6\times (\frac{1}{6}) ^3=\frac{1}{36}$$

c) Binomial distribution $n =5, K=0:1, p=\frac{1}{6}$: $ $P (\text{six spots appears at most once among the first 5 rolls}) $$ $$=\sum_{k=0}^{1}c_{5}^{k}\cdot (\frac{1}{6}) ^{k}\cdot (\frac{5}{6}) ^{5-k}\doteq0.8037551$$ R code:

> Sum (dbinom (x = 0:1, size = 5, prob = 1/6)) [1] 0.8037551

d) $ $P (\text{six spots appears for the first time on the 5th roll}) $$ $$= (\frac{5}{6}) ^4\times (\frac{1}{6}) \doteq0.0803755 1$$ R Code:

> dgeom (x = 4, prob = 1/6) [1] 0.08037551

e) There is one roll appearing six spots among the first 4 rolls and the 5th roll are six spots, that's, it is binomial di Stribution that $n =4, K=1, p=1/6$ among the first 4 rolls: $ $P (\text{six spots appears for the second time on the 5th roll }) $$ $$=c_{4}^{1}\times\frac{1}{6}\times (\frac{5}{6}) ^3\times\frac{1}{6}\doteq0.06430041$$ R code:

> dbinom (x = 1, size = 4, prob = 1/6) * 1/6[1] 0.06430041

Problem 7

I toss a coin 2 times; Then my friend tosses it 2 times. Find the chance that we get the same number of heads.

Solution:

The outcome of one is: $\{hh, HT, TH, tt\}$. Hence $ $P (\text{zero head}) =\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}$$ $ $P (\text{1 head}) =\frac{1}{2}\times\frac{ 1}{2}=\frac{1}{4}$$ $ $P (\text{2 heads}) =\frac{1}{4}\times{1}{4}=\frac{1}{16}$$ $ $P (\text{we get the same number of Heads}) =\frac{1}{16}+\frac{1}{4}+\frac{1}{16}=\frac{3}{8}$$

Problem 8

A standard deck consists of cards; + is red and is black; There is 4 cards of each rank:ace, 2, 3, 4, 5, 6, 7, 8, 9, ten, Jack, Queen, King. Cards is dealt at random without replacement. Find the chance that

A) The first 3 cards is of three different ranks

b) At least one of the first three cards is a king

c) Either the first card is a king or the second are a queen, but not both

D) There is at most one king among the first 5 cards

e) The 10th card is a king

f) The 19th and 27th cards are both kings

g) It takes more than cards for the first king to appear

h) It takes more than cards for the second king to appear

i) it takes exactly cards for the second king to appear

Solution:

A) $ $P (\text{first 3 cards is different ranks}) =\frac{52}{52}\times\frac{48}{51}\times\frac{44}{50}\doteq0.8282353$ $

b) $ $P (\text{at least one king among the first 3 cards}) $$ $$=1-p (\text{no king among the first 3 cards}) =1-\frac{48}{52}\ times\frac{47}{51}\times\frac{46}{50}\doteq0.2173756$$

C) $ $P (\text{either the 1st is king or the 2nd is Queen and not both}) $$ $$=p (\text{1st are K, 2nd is not Q}) +p (\text{1s T is isn't K, 2nd is Q}) $$ $$=p (\text{1st are K, 2nd is not Q}) +p (\text{1st are Q, 2nd is q}) +p (\text{1st are neither K nor Q, 2nd is Q}) $$ $$=\frac{4}{52}\times\frac{47}{51}+\frac{4}{52}\times\frac{3}{51}+\frac{44}{52}\times\frac{4}{51}\ doteq0.1417798$$ alternatively $ $P (\text{either the 1st is king or the 2nd is Queen and not both}) $$ $$=p (\{\text{1st are K }\}\cup\{\text{2nd is q}\})-P (\{\text{1st is k}\}\cap\{\text{2nd is q}\}) $$ $$=p (\text{1st are K}) +p (\text{2nd is Q}) -2\ti Mes P (\{\text{1st is k}\}\cap\{\text{2nd is q}\}) $$ $$=\frac{4}{52}+\frac{4}{52}-2\times\frac{4}{52}\times{4}{51}\ doteq0.1417798$$ Note that $ $P (\text{2nd are Q}) =p (\text{1st is no q, 2nd is q}) +p (\text{1st are Q, 2nd is q}) $$ $$=\frac{4} {52}\times\frac{3}{51}+\frac{48}{52}\times\frac{4}{51}=\frac{4}{52}$$

d) hypergeometric distribution $x =0:1, m=4, n=48, k=5$: $ $P (\text{at most 1 king among the first 5 cards}) $$ $$=\frac{\sum _{x=0}^{1}c_{4}^{x}\cdot c_{48}^{5-x}}{c_{52}^{5}}\doteq0.9583156$$ R Code:

> Sum (dhyper (x = 0:1, M = 4, n = x, k = 5)) [1] 0.9583156

E) $ $P (\text{10th is K}) =\frac{4}{52}$$

f) $ $P (\text{19th & 27th is both K}) =\frac{4}{52}\times\frac{3}{51}\doteq0.004524887$$

g) $ $P (\text{more than cards for the 1st K to appear}) =p (\text{no K among the first cards}) =\frac{c_{4}^{0}\cdot c_{ 48}^{10}}{c_{52}^{10}}\doteq0.4134454$$ This is hypergeometric distribution, R code:

> dhyper (x = 0, M = 4, n = x, k = 10) [1] 0.4134454

h) There might be 0 or 1 king among the first ten cards. hypergeometric distribution $x =0:1, m=4, n=48, k=10$: $ $P (\text{more than cards for the 2nd K to appear}) =p (\text{at mo St 1 K among the first cards}) $$ $$=\frac{\sum_{x=0}^{1}c_{4}^{x}\cdot c_{48}^{10-x}}{c_{52}^{10}}\doteq0.8374919$$ R Code

> Sum (dhyper (x = 0:1, M = 4, n = x, k = 10)) [1] 0.8374919

i) $ $P (\text{exactly cards for the 2nd K to appear}) $$ $$=p (\text{1 K among the first 9 cards & the 10th is K}) $$ $ $=\frac{c_{4}^{1}\cdot c_{48}^{8}}{c_{52}^{9}}\times\frac{3}{43}\doteq0.02862314$$ R Code:

> dhyper (x = 1, M = 4, n = x, k = 9) * 3/43[1] 0.02862314

The midterm

Problem 1

According to genetic theory, every plant of a particular species have a 25% chance of being red-flowering, independently of All other plants. Among plants of this species, what's the chance that fewer than 4 be red-flowering?

Solution:

Binomial distribution $n =10, K=0:3, p=0.25$: $ $P (\text{fewer than 4 is red-flowering}) =\sum_{k=0}^{3}c_{10}^{k}\cdot { 0.25}^{k}\cdot{0.75}^{10-k}\doteq0.7758751$$ R Code:

> Sum (dbinom (x = 0:3, size = ten, prob = 0.25)) [1] 0.7758751

Problem 2

A population consists of men and women. A simple random sample (draws at the random without replacement) of 4 people is chosen. Find the chance in the sample:

2 A all the people is of the same gender

2 B There is more women than men

2 c The fourth person is a woman

The third person was a woman, given that first person and fourth person was both men

Solution:

2 A $ $P (\text{all people is same gender}) =\frac{2\times c_{25}^{4}}{c_{50}^{4}}\doteq0.1098567$$

2 B $ $P (\text{more women than men}) =p (\text{4 Women 0 Man}) +p (\text{3 Women 1 Man}) $$ $$=\frac{c_{25}^{4}\times c_{25}^{0}+ C_{25}^{3}\times c_{25}^{1}}{c_{50}^{4}}\doteq0.3046027$$

2 C $ $P (\text{fourth person is woman}) =\frac{25}{50}=0.5$$

$ $P (\text{third person is woman | First and fourth were men}) =\frac{25}{48}\doteq0.5208333$$

Problem 3

An EdX multiple choice question have 5 available options, only 1 of the which is correct. Students is allowed 2 attempts at the answer. A student who does isn't know the answer decides to guess at random, as Follows:on the first attempt, he guesses at random Among the 5 options. If His guess was right, he stops. If His guess was wrong, then on the second attempt he guesses at random from among the 4 remaining options.

3 A Find the chance that the student gets the right answer on his first attempt.

3 B Find the chance that the student have to make both attempts and gets the right answer the second time.

3 C Find the chance that the student gets the right answer.

Solution:

3 A $ $P (\text{gets the right answer at first attempt}) =\frac{1}{5}=0.2$$

3 B $ $P (\text{gets the right answer the second time}) $$ $$=p (\text{second time correct | First time incorrect}) =\frac{4}{5} \times\frac{1}{4}=0.2$$

3 C $ $P (\text{gets the right answer}) $$ $$=p (\text{right answer first time}) +p (\text{right answer second time}) =0.2+0.2=0. 4$$

Problem 4

On a true-false test, each question have exactly one correct answer:true, or false. A student knows the correct answer to 70% of the questions in the test, and gives the correct answer to each of the those ques tions. Each of the remaining answers she guesses at random, independently of any other answers. After the test have been graded, one of the questions is picked at random. Given that she got the answer right and what's the chance that she knew the answer?

Solution:

According to Bayesian theorem, we have $ $P (\text{knew the answer | correct}) $$ $$=\frac{p (\text{knew the correct answer})}{p ( \text{correct})}$$ $$=\frac{p (\text{knew the correct answer})}{p (\text{knew the correct answer}) +p (\text{guessed Correct answer})}$$ $$=\frac{0.7}{0.7+0.3\times0.5}\doteq0.8235294$$

Problem 5

I Roll a die repeatedly. Find the chance that's the first 4 rolls all show different faces, and the 5th roll shows a face of that have appeared before.

Solution: $ $P (\text{first 4 Rolls different & the 5th has appeared before}) $$ $$=\frac{6}{6}\times\frac{5}{6 }\times\frac{4}{6}\times\frac{3}{6}\times\frac{4}{6}\doteq0.1851852$$

Problem 6

I throw darts repeatedly. Assume that on each single throw, my chance of hitting the bullseye are 10%, independently of all other throws. I decide to throw until I had hit the Bullseye 3 times. What's the chance that I throw exactly?

Solution:

The first throws follow the binomial distribution that $n =29, k=2, p=0.1$: $ $P (\text{hit the Bullseye 3 times among Exa CT throws}) $$ $$=p (\text{hit 2 times among the first throws and hits the bullseye on the 30th throw}) $$ $$=c_{29}^{2} \times{0.1}^2\times{0.9}^{27}\times0.1\doteq0.02360879$$ R Code:

> dbinom (x = 2, size = Prob = 0.1) * 0.1[1] 0.02360879

Problem 7

A deck of Cards consists of 8 blue cards and 5 white cards. A simple random sample (random draws without replacement) of 6 cards is selected. What's the chance that one of the colors appears twice as many times as the other?

Solution:

Follow the hypergeometric distribution that $x = 2,4, m=8, N=5, k=6$: $ $P (\text{one colors appears twice as many times as The other}) $$ $$=p (\text{4 Blue Cards & 2 White Cards}) +p (\text{2 Blue Cards & 4 White Cards}) $$ $$=\frac{c_{8}^{4 }\times c_{5}^{2}+c_{8}^{2}\times c_{5}^{4}}{c_{13}^{6}}\doteq0.4895105$$ R Code:

> Sum (dhyper (x = C (2, 4), M = 8, n = 5, k = 6)) [1] 0.4895105

Problem 8

A class has 3 teaching assistants (TAs). Each TA tosses a coin times. Find the chance that at least one of the TAs gets exactly 5 heads.

Solution:

There is processes of the binomial distribution. $ $P (\text{at least one of the Tas gets exactly 5 heads among tosses}) $$ $$=1-p (\text{none of the Tas gets exactly 5 hea DS among tosses}) $$ $$=1-c_{3}^{0}\cdot p^0\cdot{(1-p)}^{3}\doteq0.5714988$$ where $p =c_{10}^{5}\times0.5^5\ times0.5^5$. R Code:

> P = dbinom (5, ten, 1) > 1-dbinom (0, 3, p) [] 0.5714988

The University of California, Berkeley, stat2.2x probability the probability of a preliminary study note: midterm