Free math worksheets

- Decide whether each statement refers to all surveyed customers or only to a subgroup. - Words such as “both” indicate an intersection. - Phrases such as “among customers who” indicate a condition. - In \(P(A\mid B)\), the event after the vertical bar is the condition.

1. The question asks for the probability that both events occur, so the expression is \(P(P\cap D)\). 2. Buying a drink is the condition, and buying popcorn is the event of interest, so the expression is \(P(P\mid D)\). 3. Buying popcorn is the condition, and buying a drink is the event of interest, so the expression is \(P(D\mid P)\).

1. \(P(P\cap D)\) 2. \(P(P\mid D)\) 3. \(P(D\mid P)\)

- Look at the event written after the vertical bar. - Conditional probability changes the reference group rather than the event being counted.

1. The event after the vertical bar is the condition, so the restricted group is \(B\). 2. \(P(A\mid B)\) is the probability that \(A\) occurs among outcomes for which \(B\) has occurred.

The restricted group is \(B\). \(P(A\mid B)\) is the probability of \(A\) given that \(B\) has occurred.

- Use only the members in the group named after the vertical bar. - Within that group, count how many also satisfy the first event.

1. Conditioning on \(C\) means the reference group is the \(12\) chess-club members. 2. Within that group, \(3\) are also in \(R\), so \(P(R\mid C)=\frac{3}{12}=\frac{1}{4}\).

\(P(R\mid C)=\frac{1}{4}=0.25\).

- Use \(P(A\mid B)=\frac{P(A\cap B)}{P(B)}\). - For the interpretation, identify the group named by the condition and the portion of that group being measured. - In part c, pay attention to which event is the condition.

1. Use the definition of conditional probability: \(P(C\mid I)=\frac{P(I\cap C)}{P(I)}=\frac{0.18}{0.45}=0.40\). 2. This means that \(40\%\) of the students who play an instrument also sing in the choir. 3. Reverse the condition for part c: \(P(I\mid C)=\frac{P(I\cap C)}{P(C)}=\frac{0.18}{0.30}=0.60\).

a) \(P(C\mid I)=0.40\) b) Of the students who play an instrument, \(40\%\) also sing in the choir. c) \(P(I\mid C)=0.60\), or \(60\%\).

- Divide each child count by its parent count. - The probabilities on branches from the same node must add to \(1\).

1. At the first stage, \(P(A)=\frac{300}{500}=0.60\) and \(P(A^c)=\frac{200}{500}=0.40\). 2. After \(A\), \(P(B\mid A)=\frac{180}{300}=0.60\) and \(P(B^c\mid A)=\frac{120}{300}=0.40\). 3. After \(A^c\), \(P(B\mid A^c)=\frac{50}{200}=0.25\) and \(P(B^c\mid A^c)=\frac{150}{200}=0.75\).

First stage: \(0.60\), \(0.40\) After \(A\): \(0.60\), \(0.40\) After \(A^c\): \(0.25\), \(0.75\)

- For each part, identify the circle named after the vertical bar. - The overlap supplies the favorable outcomes in both directions. - Compare the sizes of the two conditioned groups.

1. Event \(B\) contains \(6+14=20\) outcomes, of which \(6\) are also in \(A\). Thus, \(P(A\mid B)=\frac{6}{20}=0.30\). 2. Event \(A\) contains \(10+6=16\) outcomes, of which \(6\) are also in \(B\). Thus, \(P(B\mid A)=\frac{6}{16}=0.375\). 3. The numerator is the same overlap count, but the conditioned groups have different sizes, so the denominators differ.

a) \(P(A\mid B)=0.30\) b) \(P(B\mid A)=0.375\) c) Both use the overlap count \(6\), but \(B\) has \(20\) outcomes while \(A\) has \(16\).

- What does a conditional probability of \(1\) tell you about the relationship between two events? - How are a joint probability, a conditional probability, and a marginal probability related? - Look for a formula that contains three of the given or unknown quantities.

1. Use the multiplication rule: \(P(S\cap F)=P(F\mid S)\cdot P(S)=0.25\cdot 0.8=0.20\). 2. Since \(P(S\mid F)=\frac{P(S\cap F)}{P(F)}\), substitute the known values: \(1=\frac{0.20}{P(F)}\). Therefore, \(P(F)=0.20\). 3. A conditional probability of \(1\) means that every student who studies French also studies Spanish.

a) \(P(S\cap F)=0.20\) and \(P(F)=0.20\). b) Every student who studies French also studies Spanish.

- Use the probability of neither defect to find the probability of at least one defect. - Apply the addition rule for two events. - In part b, the scratched phones form the denominator.

1. The probability of at least one defect is \(P(D\cup S)=1-0.85=0.15\). 2. By the addition rule, \(P(D\cap S)=P(D)+P(S)-P(D\cup S)=0.12+0.08-0.15=0.05\). 3. Therefore, \(P(D\mid S)=\frac{P(D\cap S)}{P(S)}=\frac{0.05}{0.08}=0.625\).

a) \(0.05\), or \(5\%\). b) \(P(D\mid S)=0.625\), or \(62.5\%\).

- How many ordered outcomes are possible when a die is rolled twice? - List the ordered pairs whose sum is \(10\). - For a conditional probability, restrict the sample space to outcomes that satisfy the condition. - Determine systematically which outcomes satisfy “the sum is even” and “both rolls are even.”

1. There are \(6\cdot 6=36\) equally likely ordered outcomes. A sum of \(10\) occurs for \((4, 6)\), \((5, 5)\), and \((6, 4)\), so the probability is \(\frac{3}{36}=\frac{1}{12}\). 2. An even sum occurs when both rolls have the same parity, giving \(18\) outcomes. All three outcomes with sum \(10\) are included, so the conditional probability is \(\frac{3}{18}=\frac{1}{6}\). 3. If both rolls are even, the restricted sample space has \(3\cdot 3=9\) outcomes. Only \((4, 6)\) and \((6, 4)\) have sum \(10\), so the conditional probability is \(\frac{2}{9}\).

a) \(\frac{1}{12}\) b) \(\frac{1}{6}\) c) \(\frac{2}{9}\)

- Without replacement, the same label cannot appear twice. - How many ordered pairs are possible? - When is the sum of two integers odd? - Which numbers from \(1\) through \(8\) are prime? - Once the first token is fixed, how many choices remain for the second token?

1. There are \(8\cdot 7=56\) equally likely ordered outcomes. The outcomes with sum \(9\) are \((1, 8)\), \((2, 7)\), \((3, 6)\), \((4, 5)\), \((5, 4)\), \((6, 3)\), \((7, 2)\), and \((8, 1)\). Thus, the probability is \(\frac{8}{56}=\frac{1}{7}\). 2. An odd sum occurs when one number is odd and the other is even. There are \(4\cdot 4+4\cdot 4=32\) such ordered outcomes. All \(8\) outcomes with sum \(9\) satisfy the condition, so the conditional probability is \(\frac{8}{32}=\frac{1}{4}\). 3. The prime labels are \(2, 3, 5, 7\), so there are \(4\cdot 7=28\) outcomes in which the first token is prime. The favorable outcomes are \((2, 7)\), \((3, 6)\), \((5, 4)\), and \((7, 2)\). Therefore, the conditional probability is \(\frac{4}{28}=\frac{1}{7}\).

a) \(\frac{1}{7}\) b) \(\frac{1}{4}\) c) \(\frac{1}{7}\)

- In \(P(X\mid Y)\), first identify what is already known: the event after the vertical bar. - A superscript \(^c\) denotes an event's complement. - For part b, focus on the actual condition after the test result is known. - Ask what the person wants to know about the true allergy status after seeing the result.

1. \(P(T\mid A)\) is the probability of a positive test when the allergy is present. 2. \(P(T^c\mid A)\) is the probability of a negative test even though the allergy is present. 3. \(P(T^c\mid A^c)\) is the probability of a negative test when the allergy is absent. 4. \(P(A\mid T)\) is the probability that the allergy is present when the test is positive. 5. \(P(A^c\mid T)\) is the probability that the allergy is absent even though the test is positive. 6. \(P(A^c\mid T^c)\) is the probability that the allergy is absent when the test is negative. 7. For confidence in a positive result, \(P(A\mid T)\) should be large. For confidence in a negative result, \(P(A^c\mid T^c)\) should be large.

a) (1) A positive test given that the allergy is present. (2) A negative test given that the allergy is present. (3) A negative test given that the allergy is absent. (4) The allergy is present given a positive test. (5) The allergy is absent given a positive test. (6) The allergy is absent given a negative test. b) For a positive result, \(P(A\mid T)\) should be large. For a negative result, \(P(A^c\mid T^c)\) should be large.

- In \(P(X\mid Y)\), first identify the group described after words such as “among,” “given,” or “reported.” - Decide which event describes the group being restricted and place that event after the vertical bar. - For the error-rate question, distinguish an incorrect sensor report from a correct one.

1. In a conditional probability \(P(X\mid Y)\), the event after the vertical bar identifies the group being considered. 2. For a), the group is cracked bottles, and the event of interest is being reported acceptable. This gives \(P(S^c\mid C)\). 3. For b), the group is bottles without cracks, and the event of interest is being reported defective. This gives \(P(S\mid C^c)\). 4. For c), the group is bottles reported defective, and the event of interest is actually having a crack. This gives \(P(C\mid S)\). 5. For d), the group is bottles reported acceptable, and the event of interest is actually having no crack. This gives \(P(C^c\mid S^c)\). 6. For e), the group is cracked bottles, and the event of interest is being reported defective. This gives \(P(S\mid C)\). 7. For f), the group is bottles reported defective, and the event of interest is actually having no crack. This gives \(P(C^c\mid S)\). 8. The sensor's two error rates are \(P(S^c\mid C)\), the false-negative rate, and \(P(S\mid C^c)\), the false-positive rate. Both should be small. 9. \(P(C\mid S)\) asks what proportion of defect reports correspond to cracked bottles. \(P(S\mid C)\) asks what proportion of cracked bottles are reported defective. Their conditioned groups are different, so the probabilities need not be equal.

a) \(P(S^c\mid C)\) b) \(P(S\mid C^c)\) c) \(P(C\mid S)\) d) \(P(C^c\mid S^c)\) e) \(P(S\mid C)\) f) \(P(C^c\mid S)\) g) The error rates are \(P(S^c\mid C)\) and \(P(S\mid C^c)\). Both should be small. h) \(P(C\mid S)\) conditions on bottles reported defective, while \(P(S\mid C)\) conditions on cracked bottles. They answer different questions and need not be equal.

- Split the total probability of \(B\) according to whether \(A\) occurs. - Find the known contribution through \(A\). - Use the remaining contribution and \(P(A^c)\) to find the missing conditional probability.

1. The contribution to \(P(B)\) through event \(A\) is \(P(A\cap B)=P(A)P(B\mid A)=0.4\cdot 0.7=0.28\). 2. The contribution through \(A^c\) is \(0.52-0.28=0.24\). 3. Since \(P(A^c)=0.6\), \(P(B\mid A^c)=0.24\div 0.6=0.4\).

\(P(B\mid A^c)=0.4\)

- Which group remains after the condition \(B\) is imposed? - The denominator in a conditional probability is the size of the conditioned group. - Compare the denominator Taylor used with the number of bicycle commuters.

1. Conditioning on \(B\) restricts the reference group to the \(18\) bicycle commuters, not all \(60\) commuters. 2. Within that group, \(12\) leave before \(7{:}00\) a.m., so \(P(E\mid B)=\frac{12}{18}=\frac{2}{3}\). 3. Taylor used the original sample size as the denominator instead of the size of the conditioned group.

Taylor used the wrong denominator. \(P(E\mid B)=\frac{12}{18}=\frac{2}{3}\approx 0.6667\).