Part 1: Suppose that F and X are events from a common sample space with P(F) # 0 and P(X) # 0.Prove that P(X) = P(X|F)P(F) + P(X|F)P(F). Hint: Explain why P(X|F)P(F) =P(Xn F) is another way of writing the definition of conditional probability, and then usethat with the logic from the proof of Theorem 4.1.1.%3DExplain why P(F|X) = P(X|F)P(F)/P(X) is another way of stating Theorem 4.2.1 BayesTheorem.

It is given that F and X are the two events from a common sample space with, PX=0PF =0 From the…

Answered: Part 1: Suppose that F and X are events…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Let P(E) = 0.45, P(F) = 0.4, and P(F ∩ E) = 0.25. Draw a Venn diagram and find the conditional probabilities. (a) P(E | FC )(b) P(F | EC )
If we know P(A and B) = P(A) x P(B | A), we can say event A and B are independent
) Suppose that events A and B are independent. Show that p(A | B) + p(B)(1-p(A)) - P(A ∪ B) = 0.
For this problem, assume that Pr[A∪B]=0.85and Pr[A]=0.3. (1) What is Pr[B]Pr[B] if A and B are independent events? (2) What is Pr[B]Pr[B] if A and B are disjoint events?
b. Determine P(F or D)= P(F or D). (Type an integer or a decimal.) c. Find the probability that a randomly selected adult is male. P(male) = (Type an integer or a decimal.)
Use the attached image to compute the following probabilities:• P(B|A)• P(B ∩ A)• P(B0|A)• P(B0 ∩ A)
Use the appropriate formula to answer the question: • Independent Probability: P(A and B) = P(A) x P(B) • Conditional Probability: P(A n B) = P(A) x P(B|A) • Addition Rule: P(A or B) = P(A) + P(B) – P(A and B) True or False: Events A and B are independent? P(A) = 0.60 P(B) = 0.70 P(A and B) = 0.42 True False
A student goes to the library. Let event A be that the student checks out a book and event B be that the student uses a library computer. Suppose that we happen to know that P (A) = 0.40, that P (B) = 0.30, and that P (B|A) = 0.5. Compute the following (1) P(Ac)(2) P(A AND B) (3) P(A|B)
Assume that a researcher randomly selects 14 newborn babies and counts the number of girls selected, x. The probabilities corresponding to the 14 possible values of x are summarized in the given table. Answer the question using the table. Probabilities of Girls x(girls) P(x) x(girls)| P(x) |x(girls)| P(x) 0.000 0.122 10 0.061 1 0.001 0.183 11 0.022 2 0.006 7 0.210 12 0.006 3 0.022 8 0.183 13 0.001 4 0.061 9 0.122 14 0.000 Find the probability of selecting 12 or more girls. 0.001 O 0.022 0.006 O 0.007
Let P(E) = 0.4, P(F) = 0.3, and P(F ∩ E) = 0.2. Draw a Venn diagram and find the conditional probabilities. (a) P(E | FC )(b) P(F | EC )
Look at the 4 probability rules beow , an provide a REAL LIFE example for each. Also, with a Real LIfe EXAMPLE, explain how Disjoint events CAN be dependent. Rule 1. The probability P(A) of any event A satisfies 0 ≤ P(A) ≤ 1. Rule 2. If S is the sample space in a probability model, then P(S) = 1. Rule 3. Two events A and B are disjoint if they have no outcomes in common and so can never occur together. If A and B are disjoint, P(A or B) = P(A) + P(B) This is the addition rule for disjoint events. Rule 4. For any event A, P(A does not occur) = 1 − P(A)
Suppose that A and B are independent events such thatP(A)3D0.10 and P (B) 0.60. Find P (A N B) and P (A U B). (If necessary, consult a list of formulas.) (a) P(4 n B) = 0 (b) P(4 U B) = ]

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