Suppose that E and F are two events and that P(E and F 0.7 and P(E)=0.8 what is P(F|E))

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Andrew E.

asked • 01/28/15

Im taking an online stats course and I have no idea what im doing. HELP

1 Expert Answer

Andrew E.

asked • 01/28/15

I still have no idea what I am doing. Can you help me with this one. Notice it says "or" in the last part. thanks

1 Expert Answer

Last updated at Oct. 27, 2020 by

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Question 28 (Choice 2) Given that E and F are events such that P(E) = 0.8, P(F) = 0.7, P (E ∩ F) = 0.6. Find P(𝐸 ̅|¯𝐹) Now, 𝐏(𝑬 ̅│¯𝑭)=(𝑃(𝐸 ̅ ∩ 𝐹 ̅))/(𝑃(𝐹 ̅)) =(𝑃(𝐸 ∪ 𝐹)^𝐶 )/(1 − 𝑃(𝐹)) =(𝟏 − 𝑷(𝑬 ∪ 𝑭))/(𝟏 − 𝑷(𝑭)) Finding 𝑷(𝑬 ∪ 𝑭) P(E ∪ F) = P(E) + P(F) − P(E ∩ F) Putting values P(E ∪ F) = 0.8 + 0.7 − 0.6 P(E ∪ F) = 0.8 + 0.7 − 0.6 P(E ∪ F) = 0.9 Now, 𝑃(𝐸 ̅│¯𝐹) =(1 − 𝑃(𝐸 ∪ 𝐹))/(1 − 𝑃(𝐹)) =(1 − 0.9)/(1 − 0.7) =0.1/0.3 =𝟏/𝟑

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Last updated at Oct. 27, 2020 by Teachoo

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Note : This is similar to Example 5 of NCERT – Chapter 1 Class 12 Relations and Functions

Check the answer here https:// www.teachoo.com /3959/673/Example-5---R---(a--b)---2-divides-a-b--is-equivalence-relation/category/Examples/

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Question 29 Check whether the relation R in the set Z of integers defined as R = {(𝑎, 𝑏) ∶ 𝑎 + 𝑏 is "divisible by 2"} is reflexive, symmetric or transitive. Write the equivalence class containing 0 i.e. [0]. R = {(a, b) : 𝑎 + 𝑏 is "divisible by 2"} Check reflexive Since a + a = 2a & 2 divides 2a Therefore, 2 divides a + a ∴ (a, a) ∈ R, ∴ R is reflexive. Check symmetric If 2 divides a + b , then 2 divides b + a Hence, If (a, b) ∈ R, then (b, a) ∈ R ∴ R is symmetric Check transitive If 2 divides (a + b) , & 2 divides (b + c) , So, we can write a + b = 2k b + c = 2p Adding (1) & (2) (a + b) + (b + c) = 2k + 2p a + c + 2b = 2k + 2p a + c = 2k + 2p − 2b a + c = 2(k + p − b) So, 2 divides (a + c) ∴ If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R Therefore, R is transitive. Thus, R is an equivalence relation in Z. Now, Equivalence class containing 0 i.e. [0] will be all values of a where one element is 0 Now, R = {(a, b) : 𝑎 + 𝑏 is "divisible by 2"} Putting b = 0 R = {(a, 0) : 𝑎 is "divisible by 2"} So, [0] = All possible values of a = {…., −6, −4, −2, 0, 2, 4, 6, ….}