Homework 2: Probability and its Application (MTH432/532)

English solution

Step 1 — Define the events

Let \(M\) be the event that the selected person is male, \(W\) the event that the person is female, and \(C\) the event that the person is colorblind.

Since the numbers of males and females are equal,

\[ P(M)=P(W)=\frac12. \]

The given colorblindness probabilities are

\[ P(C\mid M)=0.05=\frac5{100}, \qquad P(C\mid W)=0.02=\frac2{100}. \]

Step 2 — Find the total probability of being colorblind

A colorblind person can be either a colorblind male or a colorblind female. By the law of total probability,

\[ P(C) =P(C\mid M)P(M)+P(C\mid W)P(W). \]

Therefore,

\[ P(C) =0.05\left(\frac12\right)+0.02\left(\frac12\right) =0.025+0.01 =0.035. \]

Step 3 — Apply Bayes' theorem

We want the probability that the person is male given that the person is colorblind:

\[ P(M\mid C) = \frac{P(C\mid M)P(M)}{P(C)}. \]

Substituting the values,

\[ P(M\mid C) = \frac{0.05\cdot\frac12}{0.035} = \frac{0.025}{0.035} = \frac57. \]

Final answer

\[ \boxed{P(M\mid C)=\frac57\approx0.7143} \]

Thus, given that the randomly chosen person is colorblind, the probability that the person is male is approximately \(71.43\%\).

Quick check using 100 men and 100 women

Imagine 100 men and 100 women. About 5 men and 2 women are colorblind, so among the \(5+2=7\) colorblind people, 5 are men. Therefore the probability is \(5/7\), confirming the result.

English solution

Deck facts

A standard 52-card deck has 13 denominations (ranks): \(A,2,\ldots,10,J,Q,K\). Each denomination occurs in 4 suits.

Since two cards are selected without regard to order, the total number of 2-card hands is

\[ \binom{52}{2}=1326. \]

(a) Probability that the two cards form a pair

A pair means that the two cards have the same denomination.

First choose the denomination: \(13\) choices. For that denomination, choose any 2 of its 4 suits:

\[ 13\binom42 = 13\cdot6 = 78. \]

Hence

\[ P(\text{pair}) = \frac{13\binom42}{\binom{52}{2}} = \frac{78}{1326} = \boxed{\frac1{17}}. \]

(b) Probability of a pair given that the cards have different suits

Let

\[ A=\{\text{the two cards form a pair}\}, \qquad D=\{\text{the two cards have different suits}\}. \]

We need

\[ P(A\mid D) = \frac{P(A\cap D)}{P(D)}. \]

If two cards have the same denomination, they must automatically be from different suits because a deck has only one card of each denomination in each suit. Therefore

\[ A\subseteq D \quad\Longrightarrow\quad A\cap D=A. \]

Count the hands having different suits

The number of all 2-card hands is \(\binom{52}{2}=1326\).

The number of hands whose two cards are from the same suit is

\[ 4\binom{13}{2} = 4\cdot78 = 312. \]

Hence the number having different suits is

\[ 1326-312=1014. \]

Every pair is included among these different-suit hands, so

\[ P(A\mid D) = \frac{78}{1014} = \boxed{\frac1{13}}. \]

Final answers

• (a) \(\boxed{\frac1{17}}\)
• (b) \(\boxed{\frac1{13}}\)

Alternative check for part (b)

Condition on the first card. Given that the second card must have a different suit, there are \(39\) possible cards among the remaining deck that have a different suit from the first card. Exactly \(3\) of these have the same denomination as the first card. Thus

\[ P(\text{pair}\mid\text{different suits}) = \frac3{39} = \frac1{13}. \]

English solution

Step 1 — Define events

Let

\[ M=\{\text{rain on Monday}\}, \qquad T=\{\text{rain on Tuesday}\}. \]

The data are

\[ P(M)=0.40,\qquad P(T)=0.30,\qquad P(M\cap T)=0.20. \]

We are told that it did not rain on Monday, so we want

\[ P(T\mid M^c). \]

Step 2 — Find \(P(M^c)\)

\[ P(M^c)=1-P(M)=1-0.40=0.60. \]

Step 3 — Find the probability of Tuesday rain and no Monday rain

The event \(T\) splits into two disjoint cases: it rains on both days, or it rains Tuesday but not Monday. Therefore

\[ P(T\cap M^c) = P(T)-P(T\cap M). \]

Hence

\[ P(T\cap M^c) = 0.30-0.20 = 0.10. \]

Step 4 — Apply conditional probability

\[ P(T\mid M^c) = \frac{P(T\cap M^c)}{P(M^c)} = \frac{0.10}{0.60} = \boxed{\frac16}. \]

Final answer

\[ \boxed{\frac16\approx0.1667} \]

Thus, after learning that it did not rain on Monday, the probability of rain on Tuesday is about \(16.67\%\).

Important observation

The answer is not \(30\%\). The information that Monday was dry changes the probability because Monday and Tuesday rain are not independent under the given data: \(P(M\cap T)=0.20\), whereas independence would require \(P(M)P(T)=0.12\).

English solution

Step 1 — Define the events

Let

\[ W_T=\{\text{the transferred ball is white}\}, \qquad W_D=\{\text{the ball drawn from Urn 2 is white}\}. \]

We want

\[ P(W_T\mid W_D). \]

Step 2 — Probability that the transferred ball is white or black

Urn 1 contains 2 white and 1 black ball, so

\[ P(W_T)=\frac23, \qquad P(B_T)=\frac13. \]

Step 3 — Probability of drawing white from Urn 2 in each case

Case 1: transferred ball is white.

Urn 2 originally has \(3W+5B\). After transferring a white ball it has \(4W+5B\), total 9 balls. Therefore

\[ P(W_D\mid W_T)=\frac49. \]

Case 2: transferred ball is black.

Urn 2 then has \(3W+6B\), total 9 balls. Therefore

\[ P(W_D\mid B_T)=\frac39=\frac13. \]

Step 4 — Use Bayes' theorem

\[ P(W_T\mid W_D) = \frac{P(W_D\mid W_T)P(W_T)} {P(W_D\mid W_T)P(W_T)+P(W_D\mid B_T)P(B_T)}. \]

Substitute the values:

\[ P(W_T\mid W_D) = \frac{\frac49\cdot\frac23} {\frac49\cdot\frac23+\frac13\cdot\frac13}. \]

The numerator is

\[ \frac49\cdot\frac23=\frac8{27}, \]

and the denominator is

\[ \frac8{27}+\frac19 = \frac8{27}+\frac3{27} = \frac{11}{27}. \]

Thus

\[ P(W_T\mid W_D) = \frac{\frac8{27}}{\frac{11}{27}} = \boxed{\frac8{11}}. \]

Final answer

\[ \boxed{\frac8{11}\approx0.7273} \]

English solution

Genetic notation

Let \(B\) denote the dominant black gene and \(b\) the recessive brown gene.

• \(BB\): pure black (homozygous dominant),
• \(Bb\): hybrid black (heterozygous),
• \(bb\): brown.

What do we know about the parents?

Both parents are black, but they produced a brown offspring. A brown rat must have genotype \(bb\), so it received a \(b\) gene from each parent.

Therefore each black parent must carry one recessive brown gene. Thus both parents have genotype

\[ Bb\times Bb. \]

Possible offspring from \(Bb\times Bb\)

The four equally likely gene combinations are

\[ BB,\quad Bb,\quad Bb,\quad bb. \]

Hence

\[ P(BB)=\frac14,\qquad P(Bb)=\frac12,\qquad P(bb)=\frac14. \]

(a) Probability that the given black rat is pure black

We are already told that this particular rat is black, so genotype \(bb\) is impossible. Conditional on the rat being black, the possible outcomes are

\[ BB,\quad Bb,\quad Bb. \]

Therefore

\[ P(BB\mid\text{black}) = \frac{1}{3}. \]

Equivalently, using conditional probability,

\[ P(BB\mid\text{black}) = \frac{P(BB)}{P(\text{black})} = \frac{\frac14}{\frac34} = \boxed{\frac13}. \]

(b) Five black offspring after mating with a brown rat

The brown mate has genotype \(bb\).

There are two hypotheses for the black parent:

• \(H_1\): the black rat is pure, \(BB\), with prior probability \(1/3\);
• \(H_2\): the black rat is hybrid, \(Bb\), with prior probability \(2/3\).

Let \(F\) be the event that all 5 offspring are black.

If the parent is \(BB\)

The mating is \(BB\times bb\). Every offspring is \(Bb\), hence black. Therefore

\[ P(F\mid BB)=1. \]

If the parent is \(Bb\)

The mating is \(Bb\times bb\). Each offspring has probability \(1/2\) of being black and \(1/2\) of being brown. Assuming the offspring genotypes are independent,

\[ P(F\mid Bb) = \left(\frac12\right)^5 = \frac1{32}. \]

Apply Bayes' theorem

\[ P(BB\mid F) = \frac{P(F\mid BB)P(BB)} {P(F\mid BB)P(BB)+P(F\mid Bb)P(Bb)}. \]

Substitute the values:

\[ P(BB\mid F) = \frac{1\cdot\frac13} {1\cdot\frac13+\frac1{32}\cdot\frac23}. \]

Now

\[ \frac13=\frac{16}{48}, \qquad \frac1{32}\cdot\frac23=\frac1{48}. \]

Therefore

\[ P(BB\mid F) = \frac{16/48}{17/48} = \boxed{\frac{16}{17}}. \]

Final answers

• (a) \(\boxed{\frac13}\)
• (b) \(\boxed{\frac{16}{17}}\)

Interpretation

Before observing the offspring, a black rat from \(Bb\times Bb\) has only probability \(1/3\) of being pure \(BB\). But five consecutive black offspring are much more likely if the rat is \(BB\) than if it is \(Bb\), so the posterior probability rises to \(16/17\).

English solution

Step 1 — Define the events

Let

\[ B_1=\{\text{first ball is black}\}, \qquad R_1=\{\text{first ball is red}\}, \qquad R_2=\{\text{second ball is red}\}. \]

We must show that

\[ P(B_1\mid R_2) = \frac{b}{b+r+c}. \]

Step 2 — Probability of the first draw

Initially there are \(b+r\) balls, so

\[ P(B_1)=\frac{b}{b+r}, \qquad P(R_1)=\frac{r}{b+r}. \]

Step 3 — Second draw if the first ball was black

If the first ball is black, it is returned and \(c\) additional black balls are added.

The urn then contains

\[ b+c\text{ black balls},\qquad r\text{ red balls}, \]

for a total of \(b+r+c\) balls. Hence

\[ P(R_2\mid B_1) = \frac{r}{b+r+c}. \]

Step 4 — Second draw if the first ball was red

If the first ball is red, it is returned and \(c\) additional red balls are added. The urn then contains

\[ b\text{ black balls},\qquad r+c\text{ red balls}. \]

Therefore

\[ P(R_2\mid R_1) = \frac{r+c}{b+r+c}. \]

Step 5 — Find \(P(R_2)\)

Using the law of total probability,

\[ P(R_2) = P(R_2\mid B_1)P(B_1) + P(R_2\mid R_1)P(R_1). \]

Substitute:

\[ P(R_2) = \frac{r}{b+r+c}\cdot\frac{b}{b+r} + \frac{r+c}{b+r+c}\cdot\frac{r}{b+r}. \]

Take the common denominator:

\[ P(R_2) = \frac{br+r(r+c)} {(b+r)(b+r+c)}. \]

Simplify the numerator:

\[ br+r(r+c) = r(b+r+c). \]

Thus

\[ P(R_2) = \frac{r(b+r+c)} {(b+r)(b+r+c)} = \frac{r}{b+r}. \]

Step 6 — Apply Bayes' theorem

\[ P(B_1\mid R_2) = \frac{P(R_2\mid B_1)P(B_1)} {P(R_2)}. \]

Substituting,

\[ P(B_1\mid R_2) = \frac{ \frac{r}{b+r+c}\cdot\frac{b}{b+r} }{ \frac{r}{b+r} }. \]

Cancel the common factor \(\frac{r}{b+r}\):

\[ P(B_1\mid R_2) = \boxed{\frac{b}{b+r+c}}. \]

Hence proved

\[ \boxed{ P(\text{first black}\mid\text{second red}) = \frac{b}{b+r+c} } \]

Useful observation

Notice that \(P(R_2)=r/(b+r)\), the same as the initial proportion of red balls. Although reinforcement makes the two draws dependent, the unconditional probability that the second draw is red remains equal to the initial red proportion.

English summary

• Q1: Bayes theorem with equal male/female priors: \[ P(M\mid C)=\frac57. \]
• Q2(a): Same denomination: \[ P(\text{pair})=\frac{13\binom42}{\binom{52}{2}}=\frac1{17}. \]
• Q2(b): Pair given different suits: \[ P(\text{pair}\mid\text{different suits})=\frac1{13}. \]
• Q3: Tuesday rain given no Monday rain: \[ P(T\mid M^c)=\frac{0.30-0.20}{1-0.40}=\frac16. \]
• Q4: Transferred ball was white given final white: \[ P(W_T\mid W_D)=\frac8{11}. \]
• Q5(a): Pure black rat: \[ P(BB\mid\text{black})=\frac13. \]
• Q5(b): Pure black after five black offspring: \[ P(BB\mid\text{5 black offspring})=\frac{16}{17}. \]
• Q6: Reinforced urn: \[ P(B_1\mid R_2)=\frac{b}{b+r+c}. \]

Key ideas used

• Bayes' theorem: \[ P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B)}. \]
• Conditional probability: \[ P(A\mid B)=\frac{P(A\cap B)}{P(B)}. \]
• Law of total probability.
• Combinations for unordered card selections.
• Complement events such as \(M^c\).
• Updating probabilities after observing new evidence.