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1 Basic Probability: Problem Set One Summer We have A B B P (A B) P (B) 3. We also have from the inclusion-exclusion principle that since P (A B). P (A B) P (A) + P (B) P (A B) 3 P (A B) 3 For examples of attaining each bound, let Ω {,,..., } let F P (Ω), and let P (ω) for each ω Ω. Further, let A {ω ω 9}. When B {ω ω 9}, P (A B) and for B {ω ω 4}, P (A B). In 3 each case, P (A) 3 and P (B). 4 3 For comparable bounds on P (A B), we have A A B so 3 4 Also, P (A) P (A B). since P (A B). P (A B) 3 P (A B) Let 0,, and be the events that the coin has 0,, and heads, respectively. Let H L (resp. H U ) be the events that the lower (upper) side is heads. Note that H U H L (the event ) and P (H L ) P (H U ) P (H L ) P (H L 0) P (0) + P (H L ) P () + P (H L ) P () P (H L H U ) P (H L H U ) P (H U ) P () P (H L ) / 3/ 3

2 Let H U H L be the event that the first toss is heads up and the second is heads down. P (H U H L H U ) P (H UH L H U ) P (H U ) P (H UH L ) 3/ 3 (P (H UH L 0) P (0) + P (H U H L ) P () + P (H U H L ) P ()) ( ) P (H U H L H U H U ) P (H UH L H U H U ) P (H U H U ) P () P (H U H U ) P () P (H U H L ) / / 4 We do the final question in two parts. First, we condition on what the first coin was. First, suppose the first coin was (it cannot have been 0). P (H U H U H U H U H U ) P (H U H U H U 0 H U H U ) P (0 H U H U ) Next, suppose it was. +P (H U H U H U H U H U ) P ( H U H U ) +P (H U H U H U H U H U ) P ( H U H U ) P (H U H U H U H U H U ) P (H U H U H U 0 H U H U ) P (0 H U H U ) +P (H U H U H U H U H U ) P ( H U H U ) +P (H U H U H U H U H U ) P ( H U H U )

3 We have P ( H U H U ) P (H U H L H U H U ) 4, and since P (0 H UH U ) 0, we must have P ( H U H U ) Finally, combining the results: P (H U H U H U H U H U ) P (H U H U H U H U H U ) P ( H U H U ) +P (H U H U H U H U H U ) P ( H U H U ) P ( A C) P (B) ( P (A))P (B) P (B) P (A) P (B) P (B) P (A B) P ( A C B ) By the exact same logic, replacing B with A C and A with B in the equations above, A C and B C are independent... For pairwise independence, consider A ij and A kl, possibly with i k but definitely with j l. If the i th roll comes up x and the k th roll is y (possibly with x y), then A ij is the event that the j th roll is x and A kl is the event that the l th roll is y. Since the rolls are independent, each of these events has conditional probability and the conditional probability of both is, no matter what x and y are. Thus, the probability 3 of each event is and the probability of both is. 3 For independence of all the events, note that having all the pairs be equal is equivalent to having all the dice be equal, and ( that there are six ways to get this outcome, each ) having probability, so we have P n i<j A ij. On the other hand, there are n n(n ) pairs i < j so i<j P (A ij). For n >, these are not equal to each n(n )/ other...9 Let S be the event that the sum is 7. There are 3 equally likely outcomes of the roll, and six of them ({(n, 7 n) n,,..., }) result in a sum of 7. Therefore, P (S). Choose n for n, and let T be the event first die comes up n. So P (T ). Now S T is the same as saying the pair come up (n, 7 n) which has a probability of. So we have P (S T ) P (S) P (T ) or, so the events S and T are 3 3 independent..8. (a) Define the events: A turns up on first die and A turns up on second die, A turns up exactly once. 3

4 P (A) P ((A A ) (A A )) P (A A ) P (A A ) P (A ) + P (A ) P (A A ) (b) A first number is odd, A second number is odd. The rolls are independent, so A and A are. P (A A ) P (A ) P (A ) 4 (c) There are three ways for this to happen: A {(, 3), (, ), (3, )}. Each way is equally as likely as every other of the 3 joint possibilities, so P (A). (d) Of the numbers through, two each are congruent to 0,, and modulo 3. If the first roll is x modulo 3, then regardless of x there is a one-third chance that the second is congruent to x modulo 3 (i.e. that their sum is 0 modulo 3). Thus the probability that the sum is divisible by 3 is Let P(S) denote the power set of a set S. When F P(Ω) it is sufficient to define P (ω) for each ω Ω. (a) Let p be the probability of heads on a given flip. Let h(ω) for ω Ω be the number of total heads (e.g. h(hht ) ). Ω {HHH, HHT, HT H, HT T, T HH, T HT, T T H, T T T } F P(Ω) P (ω) p h(ω) ( p) 3 h(ω) (b) Let d(ω) be the indicator that the first and second balls have different colors ( if different, 0 o.w.). The color of the first ball is one half probability either way, and for the second ball to have the same color as the first has probability 3. Ω {UU, UV, V U, V V } F P(Ω) P (ω) + d(ω) 4

5 (c) The probability of heads coming up first on toss n (ω n) is the probability of n tails followed by a heads, in other words ( p) n p. The probability of heads never coming up (ω, countable intersection of the nested events tails comes up on each of the first n tosses ) is the limit of the probabilities of those events, lim n ( p) n, which is 0 if p > 0 and otherwise. Ω N F P (Ω) and P as defined above..8. Let D i be the event that the i th die comes up. (a) If S 4 then N < 4 so our sum in the following stops at N 4. P (N S 4) P (S 4 N ) P (N ) 4 i P (S 4 N i) P (N i) (b) Again, we must have N < 4. Let E be the event that N is even. (c) P (S 4 E) P (S 4 E) P (E) P (S 4 N ) P (N ) + P (S 4 N 4) P (N 4) i P (N i) 4 4 i ( (/4)i ) P (N S 4 D ) P (S 4 D N ) P (N ) 4 i P (S 4 D N i) P (N i)

6 (d) Define: ( N ) f(r) P D i r i ( r ) n n n n ( r ) n r/ r/ r r Note f(). Then the probability that the highest roll is exactly r is f(r) f(r ), or ( r)(3 r).8. Model this problem as randomly ordering the cherries from to 0, letting the first be those that the pig eats, and the sixteenth be the one we pick up of the remaining. (a) This is the same thing as picking one of the 0 cherries at random - the fact that five of them happen to be in the pig s stomach is not relevant. So the answer is /0, or.. (b) By the above, there are 0!/4 total orderings of the cherries having a stone in place. There are!! total orderings having all stones in places through 4 0. Thus the probability the pig ate no stones given cherry is a stone is ( 0 ), or The probability he did eat a stone given the cherry we pick is a stone, then is Inductive proof that the probability no students have the same birthday is q(m) 3!. q(), so the base case works. For the inductive step, assume the (3 m)!3 m formula works for m. Then there are 3 equally likely choices for the m th student who enters the room, of which 3 (m ) do not intersect with the set of already represented birthdays. Thus the new probability of no common birthdays is q(m ) 3 (m ) 3! 3! q(m). 3 ((3 (m ))!/(3 (m ))3 m 3 (3 m)!3 m Thus the formula works for all m, and it follows that the probability of the complement of this event, that at least students have the same birthday, is q(m). In particular, for m 3, q(m).493 and q(m).07 > There are ( ), 98, 90 total hands of poker, so to compute the probability of a given hand type, we simply divide the number of ways to get that hand type by,98,90. First consider the hands other than flushes and straights. We first choose the duplicated values (i.e.,3,...,a), then the unduplicated values, then the suits of each type. Pair: ( 3 )( 3 ) ways p.43

7 Two Pair: ( 3 )( ) ( 4 ) 3 ways p.047 Three of a Kind: ( 3 )( 3 ) 49 ways p.0 Full House: ( 3 )( 3 ) 3744 ways p.004 Four of a Kind: ( 3 )( 4 ) 4 ways p.0004 For a straight flush, we must first choose the suit (4 ways) then choose the card values (A,, 3,..., 0) that begins the straight. This gives 40 possible hands, so p For a regular flush, we choose the suit (4 ways) and then the values ( ( 3 ) ways), giving 48, then subtract off the 40 straight flushes, giving 08 ways, so p For a regular straight, we choose the starting value (0 ways), then the suits of each card (4 ways), then again subtract the 40 straight flushes, giving 000 ways, so p Let G be the distribution function of Y. If a 0 then Y b so G(y) is 0 for y < b, for all other values of y. If a > 0 then: If a < 0 then: G(y) P (Y y) G(y) P (Y y) P (ax + b y) ( P X y b ) a ( ) y b F a P (ax + b y) ( P X y b ) a ( P X < y b ) a ( lim P X y b ) + ɛ ɛ 0 a ( ) y b lim F + ɛ ɛ 0 a.7.7 The number of passengers not showing up is T Bin(0, ) for Teeny Weeny and 0 B Bin(0, ) for Blockbuster. Teeny Weeny is overbooked if T 0 and Blockbuster is overbooked if B. P (T 0) ( ) (.349 and P (B ) 9 0 ( 9 0 ) , so Teeny Weeny is overbooked slightly less often. 0 0) The number of heads out of six tosses of a fair coin is Bin(, ), so the probability of or more heads is ( ( ) + ).09. 7

8 3..3 This is identical to the situation where we throw each of n coins twice and define success as coming up heads both times. The probability of success for each coin is p and there are n such coins so the sought random variable is distributed as Bin(n, p ) and the mass function is the associated binomial mass function, f(k) ( n k) p k ( p ) n k. 8

Axioms of Probability

Axioms of Probability Sample Space (denoted by S) The set of all possible outcomes of a random experiment is called the Sample Space of the experiment, and is denoted by S. Example 1.10 If the experiment consists of tossing