Questions and Answers #4 Probability

Questions and Answers Sheet 4 Probability Question #1 What does it mean if 𝑃(𝐴|𝐵) = 𝑃(𝐴) and 𝑃(𝐵|𝐴) − 𝑃(𝐵) ? Answer: 𝐴 𝑃 ( ) = 𝑃(𝐴) 𝐵 𝑃(𝐴∩𝐵) 𝑃(𝐵) = 𝑃(𝐴) 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) ⋅ 𝑃(𝐵) This is the definition of independence of events A and B. Question #2 𝐵 Assume that A and B are events in a sample space and that 𝑃(𝐴) = .40 and 𝑃 ( ) = .25. What is the 𝐴 probability that both A and B occur? Answer: Solution: 𝐵 According to a definition of 𝑃 ( ) we have: 𝐴 𝐵 𝑃(𝐵∩𝐴) 𝐴 𝑃(𝐴) 𝑃( ) = , so we can substitute the numbers: 𝑃(𝐵 ∩ 𝐴) 0.25 = 0.4 𝑃(𝐵 ∩ 𝐴) = 0.25 ⋅ 0.4 𝑃(𝐵 ∩ 𝐴) = 0.1 Question #3 If A and B are two independent random variables, is 𝑃(𝐴|𝐵) < 𝑃(𝐴)? Answer: 𝐴 For independent variables A and B 𝑃 ( ) = 𝑃(𝐴) 𝐵 Explanation: If variables A and B are independent, then 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) ⋅ 𝑃(𝐵). 𝐴 According to definition of 𝑃 ( ) 𝐵 𝐴 𝑃( ) = 𝑃(𝐴∩𝐵) 𝐵 𝑃(𝐵) 𝐴 So: 𝑃 ( ) = 𝑃(𝐴)⋅𝑃(𝐵) 𝐵 𝑃(𝐵) 𝐴 Answer: 𝑃 ( ) = 𝑃(𝐴) 𝐵 Question #4 Two events A and B are such that 𝑃(𝐴) = 0.2, 𝑃(𝐵) = 0.3, and 𝑃(𝐴 ∪ 𝐵) = 0.4. What is 𝑃(𝐴 ∩ 𝐵) ? Answer: Explanation: 𝑃𝐴 = 0.2 𝑃𝐵 = 0.3 𝑃𝐴∪𝐵 = 0.4 𝑃𝐴∪𝐵 = 𝑃𝐴 + 𝑃𝐵 − 𝑃𝐴∩𝐵 𝑃𝐴∩𝐵 = 𝑃𝐴 + 𝑃𝐵 − 𝑃𝐴∪𝐵 Therefore 𝑃𝐴∩𝐵 = 0.2 + 0.3 − 0.4 = 0.1 If the events are independent 𝑃𝐴∪𝐵 = 𝑃𝐴 ⋅ 𝑃𝐵 = 0.5 ⋅ 0.2 = 0.1 Question #5 Suppose that 𝑃(𝐴) = 0.3 and 𝑃(𝐵) = 0.25 and 𝑃(𝐴 ∩ 𝐵) = 0.1. What is 𝑃(𝐵|𝐴(complement)) ? Answer: Answer: P(B|A')=0.5 Explanation: We use the definition of conditional probability: 𝑃(𝑋|𝑌) = 𝑃(𝑋∩𝑌) 𝑃(𝑌) along with 𝑃(𝑋 ∩ 𝑌 ′ ) = 𝑃(𝑋) − 𝑝(𝑋 ∩ 𝑌) From this we get; 𝑃(𝐵|𝐴′ ) = 𝑃(𝐵∩𝐴′ ) 𝑃(𝐴′) ′) And 𝑃(𝐵 ∩ 𝐴 = 𝑃(𝐵) − 𝑃(𝐵 ∩ 𝐴) We are given that 𝑃(𝐴) = 0.3, 𝑃(𝐵) = 0.25 and𝑃(𝐴 ∩ 𝐵) = 0.1, so: 𝑃(𝐴) = 0.3 ⇒ 𝑃(𝐴′ ) = 0.7 𝑃(𝐵 ∩ 𝐴′ ) = 0.25 − 0.1 = 0.15 0.15 𝑃(𝐵|𝐴′ ) = = 0.5 0.3 Question #6 𝑃(𝐸) = 0.24 and 𝑃(𝐹) = 0.3. What is 𝑃(𝐸 ∪ 𝐹)? Answer: The probability of P or F occurring is 0.468 Explanation: The formula is: 𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵) Basically you take the probability of E+F minus the probability of both events happening. In terms of (𝐸 ∪ 𝐹): 𝑃(𝐸 ∪ 𝐹) = 𝑃(𝐸) + 𝑃(𝐹) − 𝑃(𝐸 ∩ 𝐹) Plug in the values: 𝑃(𝐸 ∪ 𝐹) = 𝑃(0.24) + 𝑃(0.3) − 𝑃(0.24 × 0.3) 𝑃(𝐸 ∪ 𝐹) = 0.24 + 0.3 − (0.072) 𝑃(𝐸 ∪ 𝐹) = 0.54 − 0.072 𝑃(𝐸 ∪ 𝐹) = 0.468 Question #7 If 𝑃(𝐴) = 0.7, 𝑃(𝐵) = 0.9 and 𝑃(𝐴 ∩ 𝐵) = 0.4. What is P(B|A)? What about P(A|B) ? Answer: 𝑃(𝐵|𝐴) = 𝑃(𝐴|𝐵) = 𝑃(𝐴∩𝐵) 𝑃(𝐴) 𝑃(𝐴∩𝐵) 𝑃(𝐵) = = 0.4 0.7 0.4 0.9 = = 4 7 4 9 Question #8 If 𝑃(𝐴) = 0.7, 𝑃(𝐵) = 0.1, and 𝑃(𝐴 ∩ 𝐵) = 0, what is 𝑃(𝐵|𝐴) ? What about P(A|B)? Answer: Both conditional probabilities are 0. Explanation: Conditional probability is calculated as: 𝐴 𝑃(𝐴∩𝐵) 𝐵 𝐵 𝑃(𝐴) 𝑃( ) = 𝐴 If 𝑃(𝐴 ∩ 𝐵) = 0 then 𝑃 ( ) is also zero. 𝐵 𝑃 ( ) can be calculated in the same way (the numerator is also zero). 𝐴 It can also be explained this way: If events A and B cannot occur at the same time (𝑃(𝐴 ∩ 𝐵) = 0) then it is also impossible that A happened under condition of B or B under condition of A. Question #9 If 𝑃(𝐴) = 0.8, 𝑃(𝐵) = 0.9, and 𝑃(𝐴 ∩ 𝐵) = 0.8, what is P(B|A)? What about P(A|B)? Answer: 𝑃(𝐵|𝐴) = 1 8 𝑃(𝐴|𝐵) = = 0.888 9 Explanation: If 𝑃(𝐴) = 𝑃(𝐴|𝐵) then 𝐴 ⊆ 𝐵 so given B it follows that A is certain. One way to look at it is to suppose there were 1000 events P(A)=0.8 tells us that 800 of those events were of type A 𝑃(𝐴 ∩ 𝐵) = 0.8 tells us that 800 of the 1000 are both A and B So all the A events are "used up" in (𝐴 ∩ 𝐵) Using the 1000 event example: P(B)=0.9 tells us 900 of the 1000 events are of type B 800 of the 900 type B events are of type A, so 800 8 𝑃(𝐴|𝐵) = = 900 9 Question #10 If 𝑃(𝑄) = 4 7 1 and 𝑃(𝑅) = , and Q and R are independent events, then what is 𝑃(𝑄 ∩ 𝑅) ? 2 Answer: If Q and R are independent, then the probability of the event that both Q and R occur is the product of the probabilities of the two events happening on their own. Independent events are two events that do not influence each other. In other words, the outcome of one event does not depend on the outcome of the other. A good simple example is if F is the event that a coin toss comes up heads and G is the event that a die roll comes up 3. Just because a coin toss comes up heads, that doesn't change the chance of us rolling a 3 on the die. 1 1 Since the probability of tossing a head is and the probability of rolling a 3 is , 2 we have the probability of getting both a heads and a 3 as 1 1 1 × , or . 2 6 12 6 Similarly, 𝑃(𝑄 ∩ 𝑅) = 𝑃(𝑄) × 𝑃(𝑅) 4 1 = × = = 7 4 14 2 7 2