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Chapter 1: Problem 29

Suppose that \(P(E)=0.6 .\) What can you say about \(P(E \mid F)\) when (a) \(E\) and \(F\) are mutually exclusive? (b) \(E \subset F ?\) (c) \(F \subset E ?\)

Short Answer

Expert verified
(a) When E and F are mutually exclusive, \(P(E|F) = 0\). (b) When E is a subset of F, \(0.6 \leq P(E|F) \leq 1\). (c) When F is a subset of E, \(P(E|F) = 1\).

Step by step solution

01

(a) Mutually Exclusive Events

When two events are mutually exclusive, it means that they cannot both happen at the same time. In other words, if E happens, F cannot happen and vice versa. Mathematically, this means that the intersection of E and F is empty, denoted as E 鈭� F = 鈭�. Therefore, P(E 鈭� F) = 0. According to the definition of conditional probability, we have: P(E|F) = \(\dfrac{P(E 鈭� F)}{P(F)}\) Since P(E 鈭� F) = 0 in this case, the conditional probability, P(E|F), will always equal 0 because the numerator of the fraction is 0: P(E|F) = \(\dfrac{0}{P(F)}\) = 0 So, when E and F are mutually exclusive, the probability of E happening given F has happened is 0.
02

(b) E is a Subset of F

If event E is a subset of event F (E 鈯� F), it means that if F happens, then E must also happen. Mathematically, this means E 鈭� F = E. Therefore, P(E 鈭� F) = P(E). Using the definition of conditional probability, P(E|F) = \(\dfrac{P(E 鈭� F)}{P(F)}\), we can substitute the values we know: P(E|F) = \(\dfrac{P(E)}{P(F)}\) Since E is a subset of F, F could be equal to or greater than E, thus P(F) 鈮� P(E). In this case, P(E) = 0.6, so 0.6 鈮� P(F) 鈮� 1. Therefore, the range of possible values for P(E|F) is: 1 鈮� P(E|F) 鈮� \(\dfrac{0.6}{1}\) = 0.6 In summary, when event E is a subset of event F, the probability of E happening given F has happened is between 0.6 and 1, inclusive.
03

(c) F is a Subset of E

If event F is a subset of event E (F 鈯� E), it means that if F happens, then E must also happen. Mathematically, this means E 鈭� F = F. Therefore, P(E 鈭� F) = P(F). Using the definition of conditional probability, P(E|F) = \(\dfrac{P(E 鈭� F)}{P(F)}\), we can plug in the values we know: P(E|F) = \(\dfrac{P(F)}{P(F)}\) Since F has some probability of happening (P(F) > 0), we can simplify the fraction to get: P(E|F) = 1 So, when event F is a subset of event E, the probability of E happening given F has happened is 1.

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Most popular questions from this chapter

Stores \(A, B\), and \(C\) have \(50,75,100\) employees, and respectively 50 , 60, and 70 percent of these are women. Resignations are equally likely among all employees, regardless of sex. One employee resigns and this is a woman. What is the probability that she works in store \(C\) ?

If the occurrence of \(B\) makes \(A\) more likely, does the occurrence of \(A\) make \(B\) more likely?

(a) A gambler has in his pocket a fair coin and a two-headed coin. He selects one of the coins at random, and when he flips it, it shows heads. What is the probability that it is the fair coin? (b) Suppose that he flips the same coin a second time and again it shows heads. Now what is the probability that it is the fair coin? (c) Suppose that he flips the same coin a third time and it shows tails. Now what is the probability that it is the fair coin?

For events \(E_{1}, E_{2}, \ldots, E_{n}\) show that \(P\left(E_{1} E_{2} \cdots E_{n}\right)=P\left(E_{1}\right) P\left(E_{2} \mid E_{1}\right) P\left(E_{3} \mid E_{1} E_{2}\right) \cdots P\left(E_{n} \mid E_{1} \ldots E_{n-1}\right)\)

Urn 1 has five white and seven black balls. Urn 2 has three white and twelve black balls. We flip a fair coin. If the outcome is heads, then a ball from urn 1 is selected, while if the outcome is tails, then a ball from urn 2 is selected, Suppose that a white ball is selected. What is the probability that the coin landed tails?
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