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Section 1.5

Section 1.5. Events A and B are called independent events if and only if. P( A  B ) = P( A ) P( B ) . P( A ) = P( A | B )  P( B ) = P( B | A ) . Events A , B , and C are called pairwise independent if and only if.

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Section 1.5

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  1. Section 1.5 Events A and B are called independent events if and only if P(A B) = P(A) P(B)  P(A) = P(A | B)  P(B) = P(B | A) Events A, B, and C are called pairwise independent if and only if Events A, B, and C are called mutually independent if and only if both of the following hold: (a) (b) Events A1 , A2 , …, An are called mutually independent if and only if the probability of any intersection of Ais is Theorem 1.5-1 in the text: If A and B are independent events, then (a) (b) (c)

  2. 1. An urn contains seven red chips labeled distinctively with the integers 1 through 7, eight blue chips labeled distinctively with the integers 1 through 8, nine white chips labeled distinctively with the integers 1 through 9. One chip is randomly selected. The following events are defined: A = the selected chip is labeled with an odd integer, B = the selected chip is blue, C = the selected chip is labeled with a "4". (a) (b) Are A and B independent events? Why or why not? We see A and B are not independent from any one of the following:  P(A) = 13/24 P(A | B) = 4/8 = 1/2  P(B) = 8/24 = 1/3 P(B | A) = 4/13  P(AB) = 4/24 = 1/6 P(A) P(B) = (13/24)(1/3) = 13/72 Are B and C independent events? Why or why not? We see B and C are independent from any one of the following: = P(B) = 8/24 = 1/3 P(B | C) = 1/3 = P(C) = 3/24 = 1/8 P(C | B) = 1/8 = P(BC) = 1/24 P(B) P(C) = (1/3)(1/8) = 1/24

  3. Section 1.5 Events A and B are called independent events if and only if P(A B) = P(A) P(B)  P(A) = P(A | B)  P(B) = P(B | A) Events A, B, and C are called pairwise independent if and only if A and B are independent, A and C are independent, B and C are independent. (See Example 1.5-4 in the text for an example of 3 sets which are pairwise independent but not mutually independent.) Events A, B, and C are called mutually independent if and only if both of the following hold: (a) (b) A, B, and C are pairwise independent, P(A B  C) = P(A) P(B) P(C) . Events A1 , A2 , …, An are called mutually independent if and only if the probability of any intersection of Ais is the product of the individual probabilities of the Ais in the intersection. Theorem 1.5-1 in the text: If A and B are independent events, then (a) (b) (c) A and B/ are independent events, A/ and B are independent events, A/ and B/ are independent events.

  4. 2. The probability that a certain type of fire cracker will not explode is 0.2. Four fire crackers labeled 1, 2, 3, and 4 are placed side by side and triggered. Find the probability that (a) (b) there is no explosion. Ai = firecracker #i does not explode for i = 1, 2, 3, 4 It is reasonable to assume that A1 , A2 , A3 , A4 are mutually independent because of the nature of this experiment. P(no explosion) = P(A1A2A3A4) = P(A1) P(A2) P(A3) P(A4) = (0.2)4 = 0.0016 there is at least one explosion. P(at least one explosion) = 1 – P(no explosion) = 1 – 0.0016 = 0.9984

  5. 3. A fair die is rolled 5 times. Find the probability that (a) (b) at least one of the rolls results in three spots facing upward. P(at least one three) = 1 – P(no threes) = 1 – (5/6)5 = 0.5981 at most two of the rolls results in three spots facing upward. P(at most two threes) = P(no threes) + P(exactly one three) + P(exactly two threes) = 5! ——– 4! 1! 5! ——– 3! 2! (5/6)5 + (1/6)(5/6)4 + (1/6)2(5/6)3 = 0.9645

  6. 4. Suppose that events A, B, and C are mutually independent. Prove that the complement of one of these events (say A) and the other two events are mutually independent. To show that A´, B, and C are mutually independent, we must show that P(A´ B) = P(A´) P(B)P(A´ C) = P(A´) P(C) P(B C) = P(B) P(C)P(A´ B C) = P(A´) P(B) P(C) Since P(A B) = P(A) P(B), we know from Theorem 1.5-1 that P(A´ B) = P(A´) P(B) . Since P(A C) = P(A) P(C), we know from Theorem 1.5-1 that P(A´ C) = P(A´) P(C) . We already have that P(B C) = P(B) P(C) . We know from Exercise 1.5-6 that A and B C are independent, and Theorem 1.5-1 tells us that A´ and B C are independent. Therefore, it follows that P(A´ B C) = P(A´) P(B C) = P(A´) P(B) P(C) .

  7. 5. Four identical pieces of gum are to be placed into 3 mail slots. (a) Suppose a mail slot is randomly chosen for one piece of gum at a time. What is the number of equally likely possible ways this can be done? 34 = 81 (b) Suppose one of the possible ways the gum can end up being distributed in the mail slots is randomly chosen. What is the number of equally likely possible ways this can be done? 6! —— = 15 4! 2! (c) Suppose a mail slot is randomly chosen for one piece of gum at a time. What is the probability that each slot will contain at least one piece of gum? 4! —— 1! 3 4 — = — 34 27

  8. (d) Suppose one of the possible ways the gum can end up being distributed in the mail slots is randomly chosen. What is the probability that each slot will contain at least one piece of gum? 3! —— 1! 2! 6! 1 —— = — 4! 2! 5 (e) Suppose a mail slot is randomly chosen for one piece of gum at a time. What is the probability that there will be at least one empty slot? 4! —— 1! 3 23 — = — 34 27 1 – (f) Suppose one of the possible ways the gum can end up being distributed in the mail slots is randomly chosen. What is the probability that there will be at least one empty slot? 3! —— 1! 2! 6! 4 —— = — 4! 2! 5 1 –

  9. (g) Suppose a mail slot is randomly chosen for one piece of gum at a time. What is the probability that there will be exactly one empty slot? (3)(24) 16 ——— = —— 34 27 (h) Suppose one of the possible ways the gum can end up being distributed in the mail slots is randomly chosen. What is the probability that that there will be exactly one empty slot? 3! ——— 2! 1! 3 3 = — 5 6! —— 4! 2!

  10. 6. Twenty-six identical pieces of gum are to be placed into 15 mail slots. Set up each calculation, but do not actually do the calculation. (a) Suppose a mail slot is randomly chosen for one piece of gum at a time. What is the number of equally likely possible ways this can be done? 1526 (b) Suppose one of the possible ways the gum can end up being distributed in the mail slots is randomly chosen. What is the number of equally likely possible ways this can be done? 40! ——— 26! 14! (c) Suppose a mail slot is randomly chosen for one piece of gum at a time. What is the probability that each slot will contain at least one piece of gum? 26! —— 11! 1511 —— 1526

  11. (d) Suppose one of the possible ways the gum can end up being distributed in the mail slots is randomly chosen. What is the probability that each slot will contain at least one piece of gum? 25! ——— 11! 14! 40! ——— 26! 14! (e) Suppose a mail slot is randomly chosen for one piece of gum at a time. What is the probability that What is the probability that there will be at least one empty slot? 26! —— 11! 1511 —— 1526 1 – (f) Suppose one of the possible ways the gum can end up being distributed in the mail slots is randomly chosen. What is the probability that there will be at least one empty slot? 25! ——— 11! 14! 40! ——— 26! 14! 1 –

  12. (g) Suppose a mail slot is randomly chosen for one piece of gum at a time. What is the probability that there will be exactly one empty slot? (15)(1426) ———— 1526 (h) Suppose one of the possible ways the gum can end up being distributed in the mail slots is randomly chosen. What is the probability that that there will be exactly one empty slot? 25! ——— 12! 13! 15 40! ——— 26! 14!

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