Introduction to Probability: Additional Practice (Set 2)

Count the conditioning event first. A $3$ appears in $6$ rolls where the first die is a $3$ and $6$ where the second is, but $(3,3)$ was counted twice.

$\lvert B \rvert = 6 + 6 - 1 = 11$

Among those $11$ rolls, which sum to $8$? Only $(3,5)$ and $(5,3)$, since $(3,3)$ sums to $6$.

$P(\text{sum} = 8 \mid \text{at least one } 3) = \frac{2}{11}$

The value $\frac{2}{5}$ answers the reversed question, the probability of a $3$ given a sum of $8$, and the two conditionals are not the same number.

Independence lets you replace $P(A \cap B)$ with the product, so write the addition rule with $P(B)$ as the unknown $b$.

$0.6 = 0.2 + b - (0.2)b$

Collect the terms in $b$ and solve.

$0.4 = 0.8b \quad \Longrightarrow \quad b = 0.5$

Answering $0.4$ treats the events as mutually exclusive, but independent events with positive probability always overlap, so the union is smaller than the sum and $P(B)$ has to be larger to compensate.

Independence gives the intersection through the product rule.

$P(A \cap B) = P(A)\,P(B) = 0.4 \times 0.5 = 0.2$

Now feed that into the addition rule.

$P(A \cup B) = 0.4 + 0.5 - 0.2 = 0.7$

Answering $0.9$ treats the events as mutually exclusive, but independent events with positive probability are forced to overlap, so their union is strictly less than the sum.

The suit is fixed in advance, so there is no choice of suit to make. Choose five of the $13$ spades out of all five-card hands.

$P(\text{five spades}) = \frac{\binom{13}{5}}{\binom{52}{5}} = \frac{1287}{2{,}598{,}960} \approx 0.000495$

Multiplying by $4$ would answer the broader question "all five of the same suit, whichever suit that turns out to be", which is four times as likely. The expression $\left(\frac{1}{4}\right)^5$ pretends the five cards are drawn independently, but each spade removed makes the next spade less likely.

Count the favourable hands. First choose which of the four suits the hand uses, then choose $5$ of the $13$ cards in that suit.

$\lvert A \rvert = 4 \cdot \binom{13}{5} = 4 \times 1287 = 5148$

Divide by the number of hands.

$P(A) = \frac{5148}{\binom{52}{5}} = \frac{5148}{2{,}598{,}960} \approx 0.00198$

Dropping the factor of $4$ answers a narrower question, the probability that all five are the same specified suit, such as five spades. The value $\left(\frac{1}{4}\right)^5$ pretends the five draws are independent, which they are not.

"Neither" is the complement of "at least one", which is the union, so start there.

$P(A \cup B) = 0.6 + 0.3 - (0.6)(0.3) = 0.9 - 0.18 = 0.72$

Now take the complement.

$P(A^{c} \cap B^{c}) = 1 - 0.72 = 0.28$

The product route confirms it, because complements of independent events are themselves independent.

$P(A^{c})\,P(B^{c}) = 0.4 \times 0.7 = 0.28$

The sample space is the ordered lists of five faces, so $\lvert S \rvert = 6^5 = 7776$. Count the complement, where every roll dodges the $6$.

$P(\text{no } 6) = \frac{5^5}{6^5} = \frac{3125}{7776}$

Now subtract.

$P(\text{at least one } 6) = 1 - \frac{3125}{7776} = \frac{4651}{7776} \approx 0.598$

Adding $\frac{1}{6}$ five times would give $\frac{5}{6}$, which overcounts every list that contains more than one $6$.

Count the rolls in each of the three qualifying sums, working in the equally likely sample space of $36$ ordered pairs.

A sum of $10$ comes from $(4,6)$, $(5,5)$, $(6,4)$, a sum of $11$ from $(5,6)$, $(6,5)$, and a sum of $12$ only from $(6,6)$.

$\lvert A \rvert = 3 + 2 + 1 = 6$

Divide.

$P(\text{sum} \ge 10) = \frac{6}{36} = \frac{1}{6}$

The value $\frac{3}{11}$ treats the eleven possible sums as equally likely, which they are not.

The intersection is a subset of each event, so it cannot be bigger than either one.

$A \cap B \subseteq B \quad \Longrightarrow \quad P(A \cap B) \le P(B) = 0.4$

That bound is reached when $B$ sits entirely inside $A$, in which case $A \cap B = B$ and $P(A \cap B) = 0.4$.

$\max P(A \cap B) = \min\bigl(P(A),\, P(B)\bigr) = 0.4$

The value $0.2$ is what independence would give, which is legal but not the largest, and $0.5$ would exceed $P(B)$, which is impossible.

Convert the rates into counts. Imagine $20{,}000$ people, of whom $\frac{20{,}000}{200} = 100$ have the condition and $19{,}900$ do not.

The test finds $96\%$ of the $100$, giving $96$ true positives. It also fires on $3\%$ of the $19{,}900$ healthy people, giving $0.03 \times 19{,}900 = 597$ false positives.

$\text{total positives} = 96 + 597 = 693$

A positive result shrinks the sample space to those $693$ people, and only $96$ of them are genuinely affected.

$P(\text{condition} \mid \text{positive}) = \frac{96}{693} = \frac{32}{231} \approx 0.139$

The number $0.96$ is $P(\text{positive} \mid \text{condition})$. Reversing a conditional probability changes its value, often drastically, because the false positives are drawn from a far larger group.

First size the conditioning event. Of the $36$ rolls, six are doubles, so the rolls with different numbers number $36 - 6 = 30$.

Now count the rolls inside that set with at least one $5$. There are $11$ rolls containing a $5$ altogether, but one of them, $(5,5)$, is a double and has been excluded.

$\lvert A \cap B \rvert = 11 - 1 = 10$

Divide inside the shrunken sample space.

$P(A \mid B) = \frac{10}{30} = \frac{1}{3}$

The value $\frac{5}{18} = \frac{10}{36}$ comes from counting the numerator correctly but forgetting to shrink the denominator, which is the most common conditioning error.

Apply the definition of independence and then the fact that a product of two positive numbers is positive.

$P(A \cap B) = P(A)\,P(B) > 0$

An event with positive probability cannot be empty, so $A \cap B$ contains at least one outcome and the two events overlap. That immediately rules out mutual exclusivity, which would force $P(A \cap B) = 0$.

The union is therefore strictly less than the sum, since $P(A \cup B) = P(A) + P(B) - P(A)P(B)$, and $P(A \mid B) = P(A) > 0$, not $0$.