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Conditional probability

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52207910
A movie-theater survey records whether a customer buys popcorn and whether the customer buys a drink. Let \(P\) be the event “the customer buys popcorn,” and let \(D\) be the event “the customer buys a drink.” Match each description with \(P(P\cap D)\), \(P(D\mid P)\), or \(P(P\mid D)\). 1. What is the probability that a randomly selected customer buys both popcorn and a drink? 2. Among customers who buy a drink, what proportion also buy popcorn? 3. What is the probability that a customer who buys popcorn also buys a drink?

Hints

- 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.

Solution

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 \(P(D\mid P)\).

Answer

1. \(P(P\cap D)\) 2. \(P(P\mid D)\) 3. \(P(D\mid P)\)
52208010
At a high school, let \(I\) be the event “a student plays an instrument,” and let \(C\) be the event “a student sings in the choir.” A survey gives \(P(I)=0.45\), \(P(C)=0.30\), and \(P(I\cap C)=0.18\). a) Find \(P(C\mid I)\). b) Interpret your answer to part a in context. c) Find the probability that a choir member also plays an instrument.

Hints

- 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.

Solution

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: \(P(I\mid C)=\frac{P(I\cap C)}{P(C)}=\frac{0.18}{0.30}=0.60\).

Answer

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 \(60\%\).
53753610
Use the tree diagram of counts to find all six branch probabilities.
Figure for problem 537536

Hints

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

Solution

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 \(P(B^c\mid A^c)=\frac{150}{200}=0.75\).

Answer

First stage: \(0.60\), \(0.40\) After \(A\): \(0.60\), \(0.40\) After \(A^c\): \(0.25\), \(0.75\)
51475810
At a high school, students enroll in world-language courses. Let event \(S\) mean “a student studies Spanish,” and let event \(F\) mean “a student studies French.” Suppose \(P(S)=0.8\), \(P(F\mid S)=0.25\), and \(P(S\mid F)=1\). a) Find \(P(S\cap F)\) and \(P(F)\). b) Interpret \(P(S\mid F)=1\) in this context.

Hints

- 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.

Solution

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.

Answer

a) \(P(S\cap F)=0.20\) and \(P(F)=0.20\). b) Every student who studies French also studies Spanish.
52208410
A smartphone quality check records two types of defects: a display defect \((D)\) and a case scratch \((S)\). The data show that \(12\%\) of phones have a display defect, \(8\%\) have a case scratch, and \(85\%\) have neither defect. a) Find the probability that a randomly selected phone has both defects. b) Given that a phone has a case scratch, find the probability that it also has a display defect.

Hints

- 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.

Solution

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, \(P(D\mid S)=\frac{P(D\cap S)}{P(S)}=\frac{0.05}{0.08}=0.625\).

Answer

a) \(0.05\), or \(5\%\). b) \(P(D\mid S)=0.625\), or \(62.5\%\).
52208510
A fair six-sided die is rolled twice. a) Find the probability that the sum is \(10\). b) Given that the sum is even, find the probability that the sum is \(10\). c) Given that both rolls are even, find the probability that the sum is \(10\).

Hints

- 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.”

Solution

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 \(\frac{2}{9}\).

Answer

a) \(\frac{1}{12}\) b) \(\frac{1}{6}\) c) \(\frac{2}{9}\)
52208610
A bag contains eight tokens labeled \(1\) through \(8\). Two tokens are drawn in order without replacement. a) Find the probability that the sum of the two numbers is \(9\). b) Given that the sum is odd, find the probability that the sum is \(9\). c) Given that the first token shows a prime number, find the probability that the sum is \(9\).

Hints

- 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?

Solution

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 \(\frac{4}{28}=\frac{1}{7}\).

Answer

a) \(\frac{1}{7}\) b) \(\frac{1}{4}\) c) \(\frac{1}{7}\)
52211510
A company is studying a screening test for an allergy. Let \(A\) be the event “the allergy is present,” and let \(T\) be the event “the screening test is positive.” a) Describe each conditional probability in context. (1) \(P(T\mid A)\) (2) \(P(\overline{T}\mid A)\) (3) \(P(\overline{T}\mid\overline{A})\) (4) \(P(A\mid T)\) (5) \(P(\overline{A}\mid T)\) (6) \(P(\overline{A}\mid\overline{T})\) b) A person has received a test result. Which probability from part a should be large for the person to have confidence in a positive result? Which should be large for confidence in a negative result? Explain.

Hints

- In \(P(X\mid Y)\), first identify what is already known: the event after the vertical bar. - A bar over an event represents its 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.

Solution

1. \(P(T\mid A)\) is the probability of a positive test when the allergy is present. 2. \(P(\overline{T}\mid A)\) is the probability of a negative test even though the allergy is present. 3. \(P(\overline{T}\mid\overline{A})\) 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. \(P(\overline{A}\mid T)\) is the probability that the allergy is absent even though the test is positive. 6. \(P(\overline{A}\mid\overline{T})\) 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, \(P(\overline{A}\mid\overline{T})\) should be large.

Answer

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, \(P(\overline{A}\mid\overline{T})\) should be large.
52211610
At a glass-bottle factory, a sensor checks each bottle for cracks. Let \(C\) be the event “the bottle has a crack,” and let \(S\) be the event “the sensor reports a defect.” a) Write the conditional probability for each description using \(P(X\mid Y)\). (1) A cracked bottle is not detected by the sensor. (2) A bottle without a crack is incorrectly reported as defective. (3) A bottle reported as defective actually has a crack. (4) A bottle reported as acceptable actually has no crack. (5) A cracked bottle is correctly detected by the sensor. (6) A bottle reported as defective actually has no crack. b) The factory has two goals: prevent cracked bottles from reaching customers and minimize the disposal of bottles that are actually good. Explain which probabilities from part a should be large or small for each goal.

Hints

- In \(P(X\mid Y)\), the event after the vertical bar is the condition. - For the disposal goal, identify the combination in which a good bottle is reported as defective. - Bottles reach customers only after the sensor reports no defect. What should be true of bottles in that group?

Solution

1. The probabilities are: (1) \(P(\overline{S}\mid C)\), (2) \(P(S\mid\overline{C})\), (3) \(P(C\mid S)\), (4) \(P(\overline{C}\mid\overline{S})\), (5) \(P(S\mid C)\), and (6) \(P(\overline{C}\mid S)\). 2. To prevent cracked bottles from reaching customers, \(P(\overline{C}\mid\overline{S})\) should be large. The false-negative probability \(P(\overline{S}\mid C)\) should be small; equivalently, the detection probability \(P(S\mid C)\) should be large. 3. To minimize disposal of good bottles, the false-alarm probability \(P(S\mid\overline{C})\) and the probability \(P(\overline{C}\mid S)\) that a bottle reported as defective is actually good should be small. Correspondingly, \(P(C\mid S)\) should be large.

Answer

a) (1) \(P(\overline{S}\mid C)\); (2) \(P(S\mid\overline{C})\); (3) \(P(C\mid S)\); (4) \(P(\overline{C}\mid\overline{S})\); (5) \(P(S\mid C)\); (6) \(P(\overline{C}\mid S)\). b) To prevent cracked bottles from being shipped, (4) should be large, (1) should be small, and equivalently (5) should be large. To minimize disposal of good bottles, (2) and (6) should be small, while (3) should be large.
53753810
Given \(P(A)=0.4\), \(P(B\mid A)=0.7\), and \(P(B)=0.52\), find \(P(B\mid A^c)\).
Figure for problem 537538

Hints

- 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.

Solution

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 \(P(A^c)=0.6\), \(P(B\mid A^c)=0.24\div 0.6=0.4\).

Answer

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

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