CONDITIONAL PROBABILITIES A THEORETICAL APPROACH 1. Probability Space Definition 1.1. A discrete probability space (, p), abbreviaate W-space consists of a finite or countably infinite set and a function p : ! [0, 1] s.t. P!2p(!) = 1. 1.1. Events and Probability. Definition 1.2. (, p) is a W-space. Subsets of are called events. Elements of are called elementary events. Two elementary events w1,w2 are equally likely if p(w1) = p(w2) For an event A the probability of A is defined by P(A) = P!2A p(!). The empty set, is the so-called impossible event and it has probability P() = 0. The basic set itself is an event, the certain event. Definition 1.3. Let be a finite set. A probability P on is called Laplace probability (and the probability space (, P) Laplace experiment), if p(!) = P({!}) = 1 || 1.2. Conditional probabilities. Theorem 1.4. For two events A and B, P(A \ B) = P(ATBc) = P(A) − P(A \ B) A = (A \ B) [ (A \ B), (A \ B) \ (A \ B) = P(A) = P(A \ B) + P(A \ B) Theorem 1.5. P(A [ B) = P(A) + P(B) − P(A \ B) 12 A THEORETICAL APPROACH A [ B = (A − B) [ (B − A) [ (A \ B), a prtition. P(A [ B) = P(A − B) + P(B − A) + P(A \ B) = P(A) − P(A \ B) + P(B) − P(B \ A) + P(A \ B) Theorem 1.6. (, p) is a W-space and B s.t. P(B) > 0. Then there is a function pB : B ! [0, 1] s.t. (B, pB) is a W-space. Also, if A B then Pw2A pB(w) = cBP(A) where cB is a constant. Define pB : B ! [0, 1] by pB(w) = p(w) P(B) (1) Xw2B pB(w) =Xw2B p(w) P(B) = 1 P(B)Xw2B p(w) = 1 (B, pB) is a W-space. (2) Xw2A pB(w) =Xw2A p(w) P(B) = 1 P(B)Xw2A p(w) = 1 P(B)P(A) = cBP(A) Corollary 1.7. (, p) is a sample space and B s.t. P(B) > 0. Then for A PB(A) = cBP(ATB) A = (A \ B) [ (A \ B), (A \ B) \ (A \ B) = PB(A) = PB(A \ B) = cBP(ATB) since (A \ B)TB = ,ATB B Definition 1.8. Let B be an event with P(B) > 0. For each event A , P(A|B) := P(A\B) P(B) . P(A|B) is called the conditional probability for A given BAlso PB(A) = P(A|B)STOCHASTICS 3 Theorem 1.9. A,B , P(B) > 0. Then (1) A B ) P(A|B) = 1. (2) B TA = ) P(A|B) = 0. (3) If Ai, i 2 N are pairwise disjoint then P(S1i=1 Ai|B) = P1i=1 P(Ai|B) (4) P(Ac|B) = 1 − P(A|B). Proof. (1) (a), (b) follow from the definition. (2) P(S1i=1 Ai|B) = 1 P(B)P((S1i=1 Ai)TB) = 1 P(B)P(S1i=1(AiTB)) = P1i=1 P(Ai TB) P(B) = P1i=1 P(Ai|B) (3) ATAc = P(A|B) + P(Ac|B) = P(ASAc|B) = P(|B) = 1. Theorem 1.10. Let (, p) be a finite probability space, and all elementary events are equally likely. Then for A,B,B 6= , P(A|B) = |ATB| |B| Theorem 1.11. If {B1, ...,Bn} is a partition of the sample space and A is any event, then P(A) = Pni=1 P(Bi)P(A/Bi) = Sni=1 Bi,BiTBj = , i 6= j A = Sni=1 ATBi P(A) = Pni=1 P(ATBi) = Pni=1 P(Bi)P(A/Bi) Example 1.12. In an urn r red and s black balls that are made either of wood or of plastic. Let p be the plastic balls and h the wooden balls. Now let A be the event that a ball drawn is red, B is the event that it is made of plastic. (1) Find P(A) (2) Find P(B) (3) Find P(ATB)4 A THEORETICAL APPROACH (4) If nrp is the number of red plastic balls, what is the probability, given that the ball drawn is a red ball? That is find P(B/A) (5) Verify the formula P(ATB) = P(A)P(B/A) (1) P(A) = r r+s (2) P(B) = p p+h. (3) P(ATB) = nrp p+h (4) P(B/A) = nrp r (5) P(B/A) = nrp/p+h r/p+h = P(ATB)/P(A) Note that r + s = p + h Example 1.13. A five-digit number is formed with the digits 0, 1, 2, 3, 4, without repetition. Find the probability that (1) the number is divisible by 4 (2) the number ends with 0 given that the number is divisible by 4. = {(w1,w2,w3,w4,w5)|wi 2 {0, 1, 2, 3, 4} ,w1 6= 0} || = 96 (1) A = {(w1,w2,w3,w4,w5) 2 |w4w5 2 {04, 12, 20, 24, 32, 40}} |A| = 6 + 4 + 6 + 4 + 4 + 6 = 30 P(A) = 30/96 = 5/16 (2) Considering A as the sample space, B = {(w1,w2,w3,w4,w5) 2 A|w5 = 0} = {(w1,w2,w3,w4,w5) 2 |w4w5 2 {20, 40}} |B| = 6 + 6 = 12 P(B/A) = 12/30 = 6/15STOCHASTICS 5 (3) Second method = {(w1,w2,w3,w4,w5)|wi 2 {0, 1, 2, 3, 4} ,w1 6= 0} || = 96 A = {(w1,w2,w3,w4,w5) 2 |w4w5 2 {04, 12, 20, 24, 32, 40}} |A| = 6 + 4 + 6 + 4 + 4 + 6 = 30 B = {(w1,w2,w3,w4,w5) 2 |w5 = 0} |B| = 4! = 24 ATB = {(w1,w2,w3,w4,w5) 2 |w4w5 2 {20, 40}} |ATB| = 12 P(ATB) = 12/96 P(B/A) = P(ATB) P(A) = 12/30 = 6/15 1.3. 8.1.3. Example 1.14. In a box with r red and s black balls, we draw two balls without replacement. Let B be the event, that the first ball is red, A the event that the second ball is red. It is assumed that we know that the ball of the first draw is red. What is the probability that the second ball is red? After the event B has occured, there are still r − 1 red and s black balls in the urn. P(A/B) = r−1 r+s−1 Example 1.15. A bag contains 40 tickets numbered 1, 2,...,40, of which four are drawn at random and arranged in ascendiin order. Find the probability of (1) the third umber being 25. (2) the last number being 30, given that the third is 25 = {(w1,w2,w3,w4)|wi 2 [40],w1 < w2 < w3 < w4} || = 40 4 (1) A = {(w1,w2,w3,w4) 2 |w3 = 25} = {(w1,w2,w3,w4) 2 |w1,w2 2 [24],w3 = 25,w4 2 {26, ..., 40}}6 A THEORETICAL APPROACH |A| = 24 2 × 15 1 P(A) = |A| || (2) B = {(w1,w2,w3,w4) 2 |w4 = 30} = {(w1,w2,w3,w4) 2 |w1,w2,w3 2 [29]} |B| = 29 3 ATB = {(w1,w2,w3,w4) 2 |w3 = 25,w4 = 30} = {(w1,w2,w3,w4) 2 |w1,w2 2 [24],w3 = 25,w4 = 30} |ATB| = 24 2 P(B/A) = P(ATB) P(A) Example 1.16. An urn contains 5 red and 10 black balls. Two of them are drawn randomly. (1) What is the probability that both are red.? (2) If the second is red, what is the probability that the first is black? = {{w1,w2} |wi 2 {r1, ..., r5, b1, ..., b10}} || = 15 2 (1) A = {{w1,w2} 2 |w1,w2 2 {r1, ..., r5}} |A| = 52 P(A) = |A|/|| (2) B = {{w1,w2} 2 |w1 2 {b1, ..., b10}} |B| = 10 1 C = {{w1,w2} 2 |w2 2 {r1, ..., r5}} |C| = 51 B TC = {{w1,w2} 2 |w1 2 {b1, ..., b10} ,w2 22 {r1, ..., r5}}STOCHASTICS 7 |B TC| = 10 1 × 51 P(B/C) = P(B TC) P(C) Example 1.17. An urn contains 5 red and 10 black balls. Eight of them are drawn randomly. (1) What is the probability that they contain 6 black balls. (2) If one ball is taken from the 8 balls, what is the probabiilit that it is red, given that the set of eight balls in the first draw contains 6 black balls. = {{w1, ...,w8} |wi 2 {r1, ..., r5, b1, ..., b10}} || = 15 8 (1) A = {{w1, ...,w8} |w1,w2 2 {r1, ..., r5} ,w3, ...,w8 2 {b1, ..., b10}} |A| = 52 × 10 6 (2) B = {{w1} |w1 2 {r1, ..., r5}} |B| = 51 ATB = {{w1, ...,w8} |w1 2 {r1, ..., r5} ,w2, ...,w8 2 {b1, ..., b10}} Theorem 1.18. Baye’s Theorem If {B1, ...,Bn} is a partition of the sample space and A is any event, then for 1 i n P(Bi/A) = P(Bi)P(A/Bi) Pni=1 P(Bi)P(A/Bi) P(ATBi) = P(A)P(Bi/A) = P(Bi)P(A/Bi) P(Bi/A) = P(Bi)P(A/Bi) P(A) But P(A) = Pni=1 P(Bi)P(A/Bi) Hence the proof.8 A THEORETICAL APPROACH Example 1.19. Three machines M1,M2,M3 produce identicca items. Of their respective output, 5%, 4%, 3% are faulty. M1 produces 25% of the total output, M2 produces 30% and M3 produces the remainder. An item selected at random is found to be faulty. What is the probability that it was produced by the machine with the highest output? A -drawing a faulty item. For i = 1, 2, 3,Bi -drawing an item produced by Mi M1 M2 M3 Sum P(Bi) 0.25 0.30 0.45 1 P(A/Bi) 0.05 0.04 0.03 P(Bi)P(A/Bi) 0.0125 0.012 0.0135 0.038 P(Bi/A) 0.0125 0.038 0.012 0.038 0.0135 0.038 The highest outcome is from M3 Required probability = 0.0135 0.038 = 0.355 Example 1.20. Three bags. First contain 1 white, 2 red, 3 green balls. Second contains 2 white, 3 red, 1 green balls. Third contains 3 white, 1 red, 2 green balls. Two ball drawn from a bag chosen at random. They are found to be 1 white, 1 red. Find the probability that the balls so drawn came from the second bag. For i = 1, 2, 3,Bi -the i-th bag is chosen. A -the two balls are white and red. P(B1) = P(B2) = P(B3) = 1/3 P(A/B1) = 0@11 1A 0@21 1A0@62 1A = 2 15 P(A/B2) = 0@21 1A 0@31 1A0@62 1A = 25STOCHASTICS 9 P(A/B3) = 0@31 1A 0@11 1A0@62 1A = 15 P(B2/A) = P(B2)P(A/B2) PP(Bi)P(A/Bi) = 6 11 E-mail address: vattamattam@gmail.com