Free math worksheets

- Inspect the denominator in the conditional-probability definition. - A conditioning event must have positive probability for the elementary ratio. - Do not assign a value merely from the intersection.

1. The ratio would require division by \(P(B)=0\). 2. Division by zero is undefined. 3. Therefore the elementary conditional probability \(P(A\mid B)\) is not defined when the conditioning event has probability zero.

No. The usual conditional probability is undefined because its denominator is \(0\).

- Conditioning means the sample space has been narrowed to a specified event. - For sequential settings, update what remains after the stated condition occurs. - Explain what knowing the condition changes about the probability question.

1. After a red token is drawn, \(4\) red tokens remain out of \(11\) total tokens. 2. Thus \(P(\text{second red}\mid\text{first red})=\frac{4}{11}\).

\(\frac{4}{11}\).

- One probability gives the size of the conditioning group. - The other gives the share inside that group. - Convert the subgroup share to a share of all orders.

1. Use the conditional multiplication relationship. 2. \(P(\text{weekend and express})=0.40\cdot0.35=0.14\).

The joint probability is \(0.14\).

- Choose a subgroup with a higher concentration of \(A\) than the full sample. - Keep the intersection no larger than either event. - Compare the subgroup rate with the marginal rate.

1. Let \(A\) occur for \(30\) of \(100\) observations, so \(P(A)=0.30\). 2. Let \(B\) contain \(20\) observations, \(15\) of which are in \(A\). 3. Then \(P(A\mid B)=\frac{15}{20}=0.75>0.30\).

For example, \(P(A)=0.30\) and \(P(A\mid B)=0.75\) when \(15\) of \(20\) observations in \(B\) also satisfy \(A\).

- Split a reported “yes” into the two disjoint ways it can occur. - One branch depends on the unknown prevalence; the other does not. - Use the known randomization probability to solve for the population rate.

1. A “yes” response occurs either from a heads flip followed by a truthful yes or from any tails flip. 2. Therefore \(0.62=0.50p+0.50\). 3. Solving gives \(0.50p=0.12\), so \(p=0.24\). 4. A particular yes response may be truthful or forced by the coin, so it does not reveal the individual’s status. Across many responses, the known coin mechanism can be removed algebraically.

The estimated truthful-yes proportion is \(p=0.24\), or \(24\%\). An individual “yes” may be truthful or forced by the coin, protecting the person’s status, while the known randomization rate allows the population proportion to be recovered.

- Intersect both nested events with the conditioning event. - Inclusion is preserved under intersection with a fixed set. - The conditional probabilities share the same positive denominator.

1. From \(A\subseteq C\), intersecting both sets with \(B\) gives \(A\cap B\subseteq C\cap B\). 2. Probability monotonicity gives \(P(A\cap B)\le P(C\cap B)\). 3. Divide both sides by the same positive number \(P(B)\) to obtain \(P(A\mid B)\le P(C\mid B)\).

\(P(A\mid B)\le P(C\mid B)\). Intersecting both events with the same conditioning event preserves inclusion, and dividing by the same positive probability preserves the inequality.

- Rearrange the conditional-probability relationship without assuming the size of the conditioning event. - Choose two valid positive probabilities for the conditioning event. - Check the ratio in each constructed model.

1. The relationship is \(P(A\cap B)=0.40P(B)\). 2. If \(P(B)=0.50\), then \(P(A\cap B)=0.20\). 3. If \(P(B)=0.20\), then \(P(A\cap B)=0.08\). 4. Both models have \(P(A\mid B)=0.40\), so the conditional probability alone does not determine the joint probability.

No. For example: - If \(P(B)=0.50\), then \(P(A\cap B)=0.20\). - If \(P(B)=0.20\), then \(P(A\cap B)=0.08\). Both produce \(P(A\mid B)=0.40\).

- Write each conditional probability as a ratio. - Clear the denominator in each equation. - Compare the joint event appearing in both definitions.

1. From \(P(A\mid B)=\frac{P(A\cap B)}{P(B)}\), multiply by \(P(B)\) to obtain \(P(A\cap B)=P(A\mid B)P(B)\). 2. From \(P(B\mid A)=\frac{P(A\cap B)}{P(A)}\), multiply by \(P(A)\) to obtain \(P(A\cap B)=P(B\mid A)P(A)\). 3. Both expressions calculate the probability of the same intersection; they only factor it using different starting events.

\(P(A\cap B)=P(A\mid B)P(B)=P(B\mid A)P(A)\). The factors differ, but both products equal the same intersection probability.

- Find each machine’s contribution to the defective items. - Add those contributions to form the conditioning group. - Compare Machine 3’s defective contribution with the total defective probability.

1. The joint defective shares are \(0.50\cdot0.01=0.005\), \(0.30\cdot0.02=0.006\), and \(0.20\cdot0.05=0.010\). 2. The total defective probability is \(0.005+0.006+0.010=0.021\). 3. \(P(\text{Machine 3}\mid\text{defective})=\frac{0.010}{0.021}=\frac{10}{21}\approx 0.476\).

\(\frac{10}{21}\approx 0.476\).

- Multiply probabilities along one fully specified path. - Each later probability is conditional on the preceding path. - Keep the event order aligned with the given conditions.

1. Follow the specified path through the stages. 2. \(P(A\cap B\cap C)=0.60\cdot0.50\cdot0.80=0.24\).

The probability is \(0.24\).

- Describe the exact sequence required for the first success to occur on the specified roll. - Find the probability of the entire stopping-time condition. - The favorable event is contained inside the conditioning event.

1. The probability that the first \(6\) occurs on roll \(2\) is \(\frac56\cdot\frac16=\frac{5}{36}\). 2. The probability that a \(6\) occurs within three rolls is \(1-(\frac56)^3=\frac{91}{216}\). 3. The conditional probability is \(\frac{\frac{5}{36}}{\frac{91}{216}}=\frac{30}{91}\).

The probability is \(\frac{30}{91}\approx 0.330\).

- Characterize parity patterns for product and sum. - Remove patterns excluded by the condition. - Count ordered die outcomes for the surviving pattern.

1. Product even excludes only odd-odd outcomes, so the conditioning set has \(36-9=27\) outcomes. 2. An even sum occurs for even-even or odd-odd. Within the conditioning set, only even-even remains, with \(9\) outcomes. 3. The conditional probability is \(\frac{9}{27}=\frac13\).

The probability is \(\frac13\).

- Convert the two parts of the odds into shares of the conditioning event. - The two requested intersections partition event \(B\). - Check that the two joint probabilities add to the given probability of \(B\).

1. Odds of \(3\) to \(7\) within \(B\) mean \(P(A\mid B)=\frac{3}{10}\) and \(P(A^c\mid B)=\frac{7}{10}\). 2. \(P(A\cap B)=P(B)P(A\mid B)=0.40\cdot0.30=0.12\). 3. \(P(A^c\cap B)=0.40\cdot0.70=0.28\).

\(P(A\cap B)=0.12\) and \(P(A^c\cap B)=0.28\).

- Express both conditionals through the same intersection. - Use the positivity of each conditioning event. - Compare the two consequences for the intersection probability.

1. \(P(A\mid B)=1\) implies \(P(A\cap B)=P(B)>0\). 2. \(P(B\mid A)=0\) implies \(P(A\cap B)=0\). 3. The same intersection cannot have both positive probability and probability \(0\). Therefore the two statements are incompatible under the positive-margin assumptions.

No. The first statement forces \(P(A\cap B)=P(B)>0\), while the second forces \(P(A\cap B)=0\).

- Split event \(A\) into its parts inside and outside \(B\). - Use the complement of the conditioning event as the denominator. - Verify that the intersection part does not exceed the complement.

1. \(P(A\cap B^c)=P(A)-P(A\cap B)=0.55-0.25=0.30\). 2. \(P(B^c)=1-0.40=0.60\). 3. \(P(A\mid B^c)=\frac{0.30}{0.60}=0.50\).

\(P(A\mid B^c)=0.50\).

- Write both reverse conditionals with their common numerator. - The positive intersection allows cancellation. - Identify what must be equal after the common factor is removed.

1. The two conditionals are \(\frac{P(A\cap B)}{P(B)}\) and \(\frac{P(A\cap B)}{P(A)}\). 2. Because the common numerator is positive, equality of the ratios is equivalent to equality of the denominators. 3. Thus the reverse conditionals are equal exactly when \(P(A)=P(B)\).

\(P(A\mid B)=P(B\mid A)\) exactly when \(P(A)=P(B)\), under the stated positivity assumptions.

- Compute each conditional probability from its own restricted sample space. - Changing the event and changing the condition are different operations. - A conditional complement must use the same conditioning event.

1. \(P(A\mid B)=\frac12\), so \(1-P(A\mid B)=\frac12\). 2. \(B^c=\{3,4\}\) and \(A^c=\{2,3,4\}\). Thus \(P(A^c\mid B^c)=\frac22=1\). 3. Since \(1\ne\frac12\), the claim is false. 4. The correct identity keeps the conditioning event fixed: \(P(A^c\mid B)=1-P(A\mid B)\).

The proposed identity is false: \(P(A^c\mid B^c)=1\), while \(1-P(A\mid B)=\frac12\). The correct identity is \(P(A^c\mid B)=1-P(A\mid B)\).

- Find the global intersection first. - Then intersect that overlap with the conditioning event. - Conditional exclusivity concerns overlap inside the restricted sample space.

1. Globally, \(A\cap C=\{3\}\), so the events are not mutually exclusive. 2. The conditioning event \(B\) excludes outcome \(3\). Thus \(A\cap C\cap B=\varnothing\), giving \(P(A\cap C\mid B)=0\). 3. Conditioning changes the relevant outcome space. An overlap outside \(B\) has no conditional probability mass.

a) No; \(A\cap C=\{3\}\). b) Yes within \(B\); \(P(A\cap C\mid B)=0\). c) The only shared outcome is excluded by the condition.

- “Exactly one” counts the two one-sided differences but excludes the intersection. - Express the symmetric difference using the two margins and the overlap. - Use the recovered intersection as the conditional numerator.

1. The symmetric-difference identity is \(P(A\triangle B)=P(A)+P(B)-2P(A\cap B)\). 2. Thus \(0.30=0.50+0.40-2x\), so \(2x=0.60\) and \(x=0.30\). 3. \(P(A\mid B)=\frac{0.30}{0.40}=0.75\).

\(P(A\cap B)=0.30\) and \(P(A\mid B)=0.75\).

- First determine the size of the conditioning union. - Rewrite the favorable event by distributing the intersection over the union. - Watch for observations counted in both favorable pieces.

1. \(|B\cup C|=70+50-20=100\). 2. \(A\cap(B\cup C)=(A\cap B)\cup(A\cap C)\). 3. Its size is \(30+25-10=45\), subtracting the triple intersection once. 4. Therefore \(P(A\mid B\cup C)=\frac{45}{100}=0.45\).

\(P(A\mid B\cup C)=0.45\).

- Use the group named after “given” as the denominator. - Distinguish observing a relationship from assigning a treatment. - Look for a third variable that could influence both events.

1. The conditional probability is \(\frac{25}{40}=0.625\). 2. The data are observational and establish association, not the effect of an intervention. 3. Hot, sunny weather can increase both ice-cream purchases and sun exposure, creating the observed association without ice cream causing sunburn.

a) \(P(\text{sunburned}\mid\text{bought ice cream})=0.625\). b) The survey shows association only. Hot, sunny weather is a plausible common cause.

- Use only respondents in the conditional denominator. - Identify which group’s smoking outcomes are unobserved. - State what relationship between response and smoking would permit generalization.

1. The respondent group is the conditioning event, so \(P(\text{smoker}\mid\text{responded})=\frac{180}{600}=0.30\). 2. Smoking status is unknown for the \(400\) nonrespondents, so the full-sample marginal rate is not determined. 3. Using \(0.30\) for the full sample would require an assumption such as response status being independent of smoking status within the sampled population.

\(P(\text{smoker}\mid\text{responded})=0.30\). The full-sample smoking probability is not identifiable without an assumption about nonrespondents; independence of response and smoking would justify using \(0.30\).

- Treat the given comparison as odds, not a probability. - Update the odds by how much more likely the evidence is under one source. - Convert the resulting odds to a probability.

1. Prior odds for Line 2 versus Line 1 are \(1\) to \(4\). 2. Multiply the odds by the likelihood ratio \(6\), giving posterior odds of \(6\) to \(4\), or \(3\) to \(2\). 3. Odds of \(3\) to \(2\) correspond to posterior probability \(\frac{3}{3+2}=\frac35=0.60\).

Posterior odds are \(3\) to \(2\), so the posterior probability of Line 2 is \(\frac35=0.60\).

- Intersect both original events with the conditioning event. - Verify that disjoint events remain disjoint after the same intersection. - Divide the ordinary addition rule by the conditioning probability.

1. Since \(A\cap B=\varnothing\), the events \(A\cap C\) and \(B\cap C\) are also disjoint. 2. \((A\cup B)\cap C=(A\cap C)\cup(B\cap C)\). 3. Additivity gives \(P((A\cup B)\cap C)=P(A\cap C)+P(B\cap C)\). 4. Divide by \(P(C)>0\) to obtain the conditional identity. 5. Independence is irrelevant; the proof uses only disjointness and the definition of conditional probability.

\(P(A\cup B\mid C)=P(A\mid C)+P(B\mid C)\). Disjointness is sufficient; independence is not required.

- The condition changes the equally likely objects from riders to same-bus rider pairs. - Larger buses contribute many more pairs than smaller buses. - Count eligible pairs by bus type before forming the conditional ratio.

1. A small bus contributes \(\binom{10}{2}=45\) same-bus rider pairs, so all small buses contribute \(10\cdot45=450\). 2. A large bus contributes \(\binom{40}{2}=780\) pairs, so all large buses contribute \(5\cdot780=3900\). 3. Under the condition that the pair shares a bus, the eligible pair count is \(450+3900=4350\). 4. The probability of a large bus is \(\frac{3900}{4350}=\frac{26}{29}\).

The conditional probability is \(\frac{26}{29}\approx 0.897\).

- First bound the intersection using only the two marginal probabilities. - Conditional probability rescales the intersection by \(P(B)\). - Describe event arrangements that attain the smallest and largest intersections.

1. The intersection satisfies \(\max(0,P(A)+P(B)-1)\le P(A\cap B)\le\min(P(A),P(B))\). 2. Thus \(0.10\le P(A\cap B)\le0.50\). 3. Divide by \(P(B)=0.50\): \(0.20\le P(A\mid B)\le1\). 4. The lower endpoint is attained when the union has probability \(1\), and the upper endpoint is attained when \(B\subseteq A\).

\(P(A\mid B)\in[0.20, 1]\), and both endpoints are possible.

- Treat conditioning on \(B\) as a new probability model. - Apply inclusion-exclusion inside that model. - The probability of a conditional union cannot exceed \(1\).

1. Within the conditional probability measure given \(B\), the union bound gives \(P(A\cup C\mid B)\le1\). 2. Conditional inclusion-exclusion gives \(P(A\cup C\mid B)=P(A\mid B)+P(C\mid B)-P(A\cap C\mid B)\). 3. Rearranging yields the stated lower bound. 4. Numerically, \(P(A\cap C\mid B)\ge0.75+0.65-1=0.40\).

The general lower bound is \(P(A\mid B)+P(C\mid B)-1\). For the given values, \(P(A\cap C\mid B)\ge0.40\).

- Decompose the conditioning event into the part inside and outside \(A\). - Use the fact that the conditional probability equals \(1\). - Distinguish an empty set from a nonempty probability-zero set.

1. \(P(A\mid B)=1\) gives \(P(A\cap B)=P(B)\). 2. Since \(B=(A\cap B)\cup(B\setminus A)\) is a disjoint union, \(P(B\setminus A)=P(B)-P(A\cap B)=0\). 3. Thus \(B\) is contained in \(A\) up to a probability-zero part. 4. Strict set inclusion need not follow if \(B\setminus A\) is nonempty but has probability \(0\).

\(P(B\setminus A)=0\). The condition implies \(B\subseteq A\) only up to probability-zero outcomes, not necessarily as a literal set inclusion.

- Build one smaller conditioning set entirely inside \(A\). - Build another smaller conditioning set entirely outside \(A\). - Compare what happens when the same mixed set is used as the larger condition.

1. Take \(B=\{1\}\) and \(C=\{1,3\}\). Then \(P(A\mid B)=1\) and \(P(A\mid C)=\frac12\). 2. Take \(D=\{3\}\) and \(E=\{1,3\}\). Then \(P(A\mid D)=0\) and \(P(A\mid E)=\frac12\). 3. Enlarging the conditioning event can either decrease or increase a conditional probability, depending on which new outcomes are added.

a) One example is \(B=\{1\}\subset C=\{1,3\}\), giving \(1>\frac12\). b) One example is \(D=\{3\}\subset E=\{1,3\}\), giving \(0<\frac12\). c) Conditional probability is not monotone in the conditioning event.

- Break the symmetric difference into the two one-sided differences. - Compare the probabilities of the conditioning events and of their intersections with \(A\). - Probability-zero discrepancies do not change either conditional ratio.

1. Since \(B\triangle C\) has probability \(0\), the differences \(B\setminus C\) and \(C\setminus B\) each have probability \(0\). 2. Therefore \(P(B)=P(B\cap C)=P(C)\). 3. The sets \(A\cap B\) and \(A\cap C\) can differ only inside \(B\triangle C\), so their probabilities are equal. 4. Equal numerators and equal positive denominators give \(P(A\mid B)=P(A\mid C)\).

For every event \(A\), \(P(A\mid B)=P(A\mid C)\). Conditioning is unchanged when the conditioning events differ only on a null set.

- Treat the condition as a restricted geometric region. - Find the area satisfying both conditions. - Divide intersection area by conditioning-region area.

1. The conditioning region is the triangle above \(y=x\), with area \(\frac12\). 2. In the half-strip \(x<\frac12\), the region above the line has area \(\frac12-\frac12\cdot\frac12\cdot\frac12=\frac38\). 3. The conditional probability is \(\frac{\frac38}{\frac12}=\frac34\).

The probability is \(\frac34\).

- Split the conditioned intersection across the two disjoint parts. - Insert each subgroup probability as a factor and its reciprocal. - Interpret the result as a weighted average within the union.

1. Since \(B\) and \(C\) are disjoint, \(A\cap(B\cup C)=(A\cap B)\cup(A\cap C)\) is a disjoint union. 2. Divide by \(P(B\cup C)\): \(P(A\mid B\cup C)=\frac{P(A\cap B)}{P(B\cup C)}+\frac{P(A\cap C)}{P(B\cup C)}\). 3. Factor each term as \(\frac{P(A\cap B)}{P(B)}\cdot\frac{P(B)}{P(B\cup C)}\) and similarly for \(C\). 4. These factors are the conditional probabilities in the stated identity.

\(P(A\mid B\cup C)=P(A\mid B)P(B\mid B\cup C)+P(A\mid C)P(C\mid B\cup C)\).

- Rewrite both conditional probabilities in the ratio using the same denominator. - Use the subset relationship to simplify intersections with \(B\). - Consider what the numerator and denominator would represent if part of \(C\) lay outside \(B\).

1. Since \(C\subseteq B\), \((A\cap C)\cap B=A\cap C\) and \(C\cap B=C\). 2. Therefore \(P(A\cap C\mid B)=\frac{P(A\cap C)}{P(B)}\) and \(P(C\mid B)=\frac{P(C)}{P(B)}\). 3. Dividing cancels \(P(B)\): \(\frac{P(A\cap C\mid B)}{P(C\mid B)}=\frac{P(A\cap C)}{P(C)}=P(A\mid C)\). 4. Without \(C\subseteq B\), conditioning on \(B\) replaces \(C\) by \(B\cap C\), so the ratio generally describes conditioning on \(B\cap C\), not on \(C\).

\(P(A\mid C)=\frac{P(A\cap C\mid B)}{P(C\mid B)}\). The subset condition ensures that conditioning first on \(B\) does not remove any outcomes from \(C\).

- Express both marginal probabilities in terms of the shared intersection. - Substitute those expressions into the union formula. - Solve for the intersection before recovering the marginals.

1. Let \(I=P(A\cap B)\). Then \(P(B)=\frac{I}{0.60}\) and \(P(A)=\frac{I}{0.40}\). 2. The union equation is \(\frac{I}{0.40}+\frac{I}{0.60}-I=0.70\). 3. This gives \(\frac{19}{6}I=0.70\), so \(I=\frac{21}{95}\), \(P(A)=\frac{21}{38}\), and \(P(B)=\frac{7}{19}\).

\(P(A\cap B)=\frac{21}{95}\), \(P(A)=\frac{21}{38}\), and \(P(B)=\frac{7}{19}\).

- Each pair of reverse conditionals shares the same intersection numerator. - Convert each reverse-conditional ratio into a ratio of marginal probabilities. - Multiply the three marginal ratios around the cycle and look for cancellation.

1. For positive intersections, \(\frac{P(A\mid B)}{P(B\mid A)}=\frac{P(A)}{P(B)}\), and similarly around the cycle. 2. Multiplying the three ratios must give \(\frac{P(A)}{P(B)}\cdot\frac{P(B)}{P(C)}\cdot\frac{P(C)}{P(A)}=1\). 3. The reported ratios give \(\frac{0.60}{0.30}\cdot\frac{0.50}{0.50}\cdot\frac{0.40}{0.40}=2\), not \(1\). 4. Therefore the claims are inconsistent.

No. The cycle of reverse-conditional ratios must multiply to \(1\), but the reported values give \(2\).

- Rewrite every conditional probability as an intersection divided by the fixed denominator. - Check what happens when the event is the full sample space. - Intersecting disjoint events with the same conditioning event preserves disjointness.

1. \(Q(A)=\frac{P(A\cap B)}{P(B)}\ge0\) because ordinary probabilities are nonnegative and \(P(B)>0\). 2. \(Q(S)=\frac{P(S\cap B)}{P(B)}=\frac{P(B)}{P(B)}=1\). 3. If \(A_1,A_2,\ldots\) are pairwise disjoint, then \(A_1\cap B,A_2\cap B,\ldots\) are also pairwise disjoint. 4. By countable additivity of \(P\), the probability of the union of the events \(A_i\cap B\) is \(\sum_i P(A_i\cap B)\). Dividing by the fixed positive value \(P(B)\) gives \(\sum_i Q(A_i)\), so \(Q\) is countably additive. 5. Thus conditioning on \(B\) creates a valid probability measure on the same event collection.

The function \(Q(A)=P(A\mid B)\) satisfies nonnegativity, normalization, and countable additivity for pairwise disjoint events, so it is a probability measure.