Joint Events, Sequential Trials, and Independence

Problem
An urn contains $10$ black and $5$ white balls. Draw two successively without replacement. Find the probability that both are black.

1. 1.Let $E$ mean the first is black and $F$ mean the second is black. Initially $P(E)=10/15$.
2. 2.Given a first black ball, $9$ black balls remain among $14$, so $P(F\mid E)=9/14$.
3. 3.Therefore $P(E\cap F)=(10/15)(9/14)=3/7$. The second denominator and the number of black balls both change because the first ball is not returned.

Problem
Three cards are drawn successively without replacement from a standard $52$-card deck. Find the probability of king, king, ace in that order.

1. 1.The first card is a king with probability $4/52$. Once that king is removed, the second is a king with conditional probability $3/51$.
2. 2.After two kings are removed, all four aces remain among $50$ cards. The final conditional probability is $4/50$.
3. 3.Multiply along the specified sequence: $(4/52)(3/51)(4/50)=2/5525$. There is no factor for rearranging the cards, because their order has been specified.

Problem
On a fair die let $E=\{3,6\}$ be a multiple of $3$ and $F=\{2,4,6\}$ be an even result. Test independence.

1. 1.The probabilities are $P(E)=2/6=1/3$ and $P(F)=3/6=1/2$.
2. 2.The overlap is $\{6\}$, so $P(E\cap F)=1/6$. The product is $(1/3)(1/2)=1/6$.
3. 3.The events are independent even though they describe the same throw and have a common outcome. Independence must be checked using probabilities, not guessed from the wording.

Problem
A fair die is thrown twice. Are an odd first result and an odd second result independent?

1. 1.Among $36$ equally likely ordered outcomes, $18$ have an odd first result and $18$ have an odd second result. Each event has probability $1/2$.
2. 2.There are $3\times 3=9$ outcomes odd in both positions, so the joint probability is $9/36=1/4$.
3. 3.Because $1/4=(1/2)(1/2)$, the two events are independent. Independent experiments require this product relationship for every pair of events belonging to the separate experiments.

Problem
Toss three fair coins. Let $E$ mean all heads or all tails, $F$ mean at least two heads, and $G$ mean at most two heads. Classify the three pairs.

1. 1.Counting the eight outcomes gives $P(E)=1/4$, $P(F)=1/2$, and $P(G)=7/8$.
2. 2.The intersections have probabilities $P(E\cap F)=1/8$, $P(E\cap G)=1/8$, and $P(F\cap G)=3/8$.
3. 3.Compare each with its product: $P(E)P(F)=1/8$, $P(E)P(G)=7/32$, and $P(F)P(G)=7/16$. Only $E,F$ are independent. Each pair needs its own check.

Problem
Show that independence of $E$ and $F$ implies independence of $E$ and $F^{\prime }$.

1. 1.Split $E$ into disjoint pieces: $E=(E\cap F)\cup (E\cap F^{\prime })$. Hence $P(E\cap F^{\prime })=P(E)-P(E\cap F)$.
2. 2.Use independence to replace the joint probability: $P(E\cap F^{\prime })=P(E)-P(E)P(F)=P(E)[1-P(F)]$.
3. 3.Since $1-P(F)=P(F^{\prime })$, we obtain $P(E\cap F^{\prime })=P(E)P(F^{\prime })$. The same reasoning shows that $E^{\prime },F$ and $E^{\prime },F^{\prime }$ are independent.

Problem
For independent $A,B$, express the probability that at least one occurs using their complements.

1. 1.The opposite of at least one is neither. In symbols, $(A\cup B{)}^{\prime }=A^{\prime }\cap B^{\prime }$.
2. 2.The complements are independent, so $P(A^{\prime }\cap B^{\prime })=P(A^{\prime })P(B^{\prime })$.
3. 3.Thus $P(A\cup B)=1-P(A^{\prime })P(B^{\prime })$. Equivalently, substitute $P(A\cap B)=P(A)P(B)$ into the ordinary addition rule and factor the result.

Problem
A and B take turns rolling a fair die until someone rolls $6$. A starts. Find their winning probabilities.

1. 1.Write success probability $p=1/6$ and failure probability $q=5/6$. A can win on throw $1,3,5,\dots$. Those disjoint possibilities have probabilities $p,q^{2}p,q^{4}p,\dots$.
2. 2.Every extra chance for A requires two further failures. Hence $P(A\text{ wins})=p[1+q^2+q^4+\cdots ]=p/(1-q^2)$.
3. 3.The geometric-series formula follows because multiplying $1+r+r^2+\cdots$ by $r$ shifts it one term, leaving $1$ after subtraction; for $|r|<1$, the remaining tail tends to zero. Here $p/(1-q^2)=(1/6)/(1-25/36)=6/11$.
4. 4.The probability of no success after $n$ throws is $q^n$, which tends to zero, so someone wins with probability one. Thus B wins with probability $1-6/11=5/11$. A has the advantage of the first opportunity.