STAT 102 - Probability and Statistics (PDF)

# STAT 102 ## Random Experiment An experiment is called a random experiment if when conducted the outcome or result cannot be predetermined. If repeated many times it is essentially a homogeneous condition, the result is not unique but may be any of the various possible outcomes. ## Terminology...

# STAT 102 ## Random Experiment An experiment is called a random experiment if when conducted the outcome or result cannot be predetermined. If repeated many times it is essentially a homogeneous condition, the result is not unique but may be any of the various possible outcomes. ## Terminology **Trial:** Performing a random experiment is called a trial. **Event:** Outcome or combination of outcomes are termed as event. For example, tossing a coin 2 times. S = {HH, HT, TI, TH} 1. (Obtaining the same outcome) = {HH, TT} 2. (Obtaining one head) = {HT, TH} 3. (Obtaining at least one head) = {HH, HT, TI, TH} 4. (Obtaining 3 heads) = {} **Exclusive Cases:** The total number of possible outcomes of RE is called the Exclusive for the experiment. **Mutually Exclusive Event:** Two or more events are mutually exclusive if the happening of any one of them excludes the happening of the others in the same time. **Exhaustive Lively Cases:** Outcomes are said to be mutually or equally possible if none of them is expected to occur in preference to show thus in tossing of a coin is unbiased. **Independent Event:** Events are said to be independent of each other if the happening of any of them is not affected by and does not affect the happening of any other. Tossing coin and die S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6} Note: H1 & H6 E1 (Obtaining H&6) = {H6} E2 (Obtaining H & even no) = {H2, H4, H6} E3 (H & any no) = {H1, H2, H3, H4, H5, H6} E4 (H& any no of H or T) = {H6, T6} If Ei and Ej are two events, then Ei & Ej are independent IF P(EiEj) = P(Ei) P(Ej). But if P(EiEj) = P(Ei) P(EiEj) or = P(Ei) P(Ej|Ei) then Ei & Ej are dependent. P(H6) = 1/2 = P(E1) = P(E3 & E4) = 1/2 ## Definition of Probability Mathematical or classical prior probability - If a random experiment results in n exclusive mutually exclusive equally likely outcome, out of which m are favorable to the happening of an event A, then the probability of occurrence of A usually denoted by P (A) is given by: P(A) = m/ favorable no of cases A/ exclusive no of cases = P(A) = N **Remark** 1. P (not event A) = 1 - P(A) = favorable cases to A/ exclusive no of cases = N - M/ N = 1 - M/ N = 1 - P(A) Since m and n are non-negative and m < n, then P (A) ≥ 0 and P (A) ≤ 1 Hence 0 ≤ P (A) ≤ 1 **The Classical (Priori) Prob: ** has its short coming and it fails in the following situations: 1. If n, the exclusive number of cases of the randome experiment is infinite 2. If the various outcomes of the random experiment are not equally likely ## Statical or Empirical Probability If an experiment is performed repeatedly under essentially homogeneous & identical conditions then the limiting value of the ratio of the number of times the event occurs to the number of identically large is called the probability of happening of the event. It is being assumed that the limit is finite. ## Example 1 If a fair die is thrown at random find the probability that the number on it is (i) five (ii) greater than 4 (iii) even number. **Solution** If a fair die is thrown, the exclusive number of cases is six and one is equally likely. S = {1, 2, 3, 4, 5, 6}, n (S) = 6 1. Let A be an event of obtaining five. A = {5}, n(A) = 1 P(A) = no cases in A/ no of cases in (S) = 1/6 2. Let B are event of obtaining greater than 4. B = {5, 6}, n(B) = 2 P(B) = n (B)/ n (S) = 2/6 = 1/3 3. Let C be an event of obtaining an even number on the die. C = {2, 4, 6}, n(C) = 3 P(C) = n(C)/ n (S) = 3/6 = 1/2 ## Example 2 Four cards are drawn at random from a pack of 52 cards. Find the probability that: 1. Two are king & two are aces 2. They are a king, a queen, a jack and an ace 3. All are diamond 4. Two are reds & two are black There is one card for each suit. Note that there are 52 cards in a pack of cards. There are 2 colors in the pack of cards. 13 are red & 26 are black if we consider diamond, spade, hearts and stars in classifying them: 13 diamond 13 spade 13 hearts 13 stars Card from each of the 13 above we have: king, queen, jack, and ace. Therefore we have: 4 kings 4 queens 4 jacks 4 aces **Solution** For cards can be drawn, the pack of cards in 52C4 ways. Which gives the exclusive number of cases. 1. P (king, queen, jacks, aces) =? One king can be drawn out of 4 kings in 4C1. Similarly one queen is 4C1, one jack 4C1 and one ace. Therefore we have 4C1, 4C1, 4C1, 4C1 favourable outcomes. P(king, queen, jacks, aces) = (4C1)^4/ 52C4 = 256/ 52C4 ## Combination & Permutation 1. Consider ABC, if 2 are taken out of ABC, we have different sets, do we have? AB, AC, BC Soln The sets are AB, AC & BC. 2. According ABCD if 2 are taken different, set as we have: Soln 4 could be AB, AC, AD, BC, BD & CD. Note: In combination order is not important, it means AB = BA 3. ABCDE out 2 are taken Soln, it could be AB, AC, AD, AE, BC, BD, BE, CD, CE, DE. A combination of n different objects taken r at a time donated by "nCr or nPr" is a solution of r objects out of n objects without any regard to the order of arrangements. Consider 4 different objects say A, B, C, D taken 2 at a time, then we have AB, AC, AD, BC, BD, CD nCr = n!/ r!(n-r)! Note that: n! = n(n-1)(n-2)...3, 2, 1 0! = 1 If n = 4 and r = 2 4C2 = 4!/ 2!(4-2)! = 4x3x2x1/ 2x1(2x1) = 12/ 2 = 6 If n = 10 and r = 4, what is 10C4. Soln 10C4 = 10!/ 4!(10-6)! = 10x9x8x7x6!/ 4x3x2x1x6! = 10x9x7/ 3 = 630/ 3 = 210 Note: nCr = n!/(n-r)! If out of 3 male and 4 female students, 2 are selected at random, how many sets of 2 students can be made if: 1. are of same gender 2. different gender and 3. only males Soln Let m1, m2, m3, f1, f2, f3, f4 represent male and female students. 1. m1, m2, m3, f1, f2, f3, f4 It could be both are male or female. m1m2 + f1f2 m1m3 + f1f3 m2m3 + f2f3 f2f4 f3f4 = 3 + 6 = 9 Then 9 different set of the same gender. Or 3C2 + 4C2 = 3 + 6 = 9 2. Are of different gender m1f1 m2f1 m3f1 m1f2 m2f2 m3f2 m1f3 m2f3 m3f3 m1f4 m2f4 m3f4 4 + 4 + 4 + 4 = 12 There are 12 different sets of 2 students that are of different gender. Or. 3C1 x 4C1 = 3 x 4 = 12 ___ 3. Only male students. It can be m1m2, m1m3, m2m3 or 3C2 = 3 There are 3 different sets of 2 students. **Irrespective of gender** It could be both male, both female or one male and female. (3 male + 4 female) 7C2 = 7!/ 2!(7-2)! = 7x6x5!/ 2x1x5! = 21 **OR** ## Permutation **Definition:** Permutation of n objects taken r at a time denoted by "nPr" is an arrangement of only r objects of the n objects generally, if we have n distinct objects, then objects can be arranged in n! ways. If we have A, B and C, these can be arranged as: ABC ACB BCA BAC CAB CBA Here n=3 The arrangements of 3 distinct objects can be done in 3! ways 3! = 3x2x1 x1 = 6 ways Assuming A, B, C and D ABCD ABDC ACBD ACDB ADBC BADC BCAD BCDA BDAC BDC A CABD CADB In how many ways that ABBA can be arranged? Soln ABBA ABAB AABB BABA BAAB BBAA Mathematically, this can be done as: 4!/ 2!2! = 4x3x2x1/ 2x1x2x1 = 12 Generally, if we have n objects n are alike N are also alike an so on.... such that n = n1 + n2 +....n1, the arrangements can be done as: n!/ n1!n2!n3!... For ABBA n=4, n1=2, n2=2 The arrangement of the word ABBA can done as: 4!/ 2!2! = 12 ways In how many ways the following word can be arranged: 1. STATISTIC 2. COMMITTEE Soln: 1. n=10, If n1 represent S = 3 n2 = T = 3 n3 = I = 2 n4 = C = 1 n5 = (E) = 1 n6 = T = 1 The word STATISTIC can be arranged in 10!/ 3!3!2!1!1!1! = 10! / 3! x 3! x 2! = 10x9x8x7x6x5x4x3x2x1 / 3!x(2x2)x(2x1) = 50,400 ways 2. In the word COMMITTEE, we have n = 9, n1(C)=2, n2(O) = 2, n3(M) = 2, n4(I) = 1, n5(T) = 1 n6(E) = 1 The word COMMITTEE can be arranged in 9!/ 2! 2! 2! 1! 1! 1! = 45360 ways ## In how many ways that 10 students can be seated on 4 seats? Soln: If 4 students out of the 10 students seated the 3 seats, then it can be done in 10P4 ways. 10P4 = 10!/ (10-4)! = 10x9x8x7x6/ 6! = 5040 ## How many ways 4 digit number can be formed from 10 digits 0, 1, 2, 3.....9 if: 1. repetition are allowed 2. repetition are not allowed 3. the last digit must be zero and repetition are not allowed? Soln The 1st digit can be any of the nine digits 1, 2, 3....9 9 x 10 x 10 x 10 = 9000 9 x 8 x 7 x 6 = Because the last one is zero. **A combination is a unordered selection of object and is all about selection** **A permutation is order arrangement of object.** ## Sample Space The set of all possible outcomes of a random experiment is known as the sample space and denoted by "S". In other words, Sample Space is the set of all exhaustive cases of a random experiment. For example, if a die is tossed once, the sample s = {1, 2, 3, 4, 5, 6}. Similarly, if 2 dice are tossed the sample space: dield2| 1 | 2 | 3 | 4 | 5 | 6 ---| ---| ---| ---| ---| ---| --- 1| 1,1 | 1,2 | 1,3 | 1,4 | 1,5 | 1,6 2 | 2,1 3 | 3,1 4 | 4,1 5 | 5,1 6 | 6,1 n(S) = 36 elements If we are interested in the sum of the outcomes, the sample space will be: dield2 | 1 | 2 | 3 | 4 | 5 | 6 ---| ---| ---| ---| ---| ---| --- 1 | 2 | 3 | 4 | 5 | 6 | 7 2 | 3 | 4 | 5 | 6 | 7 | 8 3 | 4 | 5 | 6 | 7 | 8 | 9 4 | 5 | 6 | 7 | 8 | 9 | 10 5 | 6 | 7 | 8 | 9 | 10 | 11 6 | 7 | 8 | 9 | 10 | 11 | 12 What is the probability of getting the sum is four P (event) = n (event)/ n (S) P (Sum = 4) = n (element their sum is 4)/ n (s) = 3/ 36 P (sum is 6 or 7) = P (sum is 6) + P (sum is 7) = 5/36 + 6/36 = 11/36 P (sum is 6 & 2) = 0 = 5/36 x 6/36 = 0 ## Aromantic Probability Given a sample space of a random experiment. The probability of the occurrence of any event “A' donated by P(A) satisfying the following axioms 1. P(A) is defined real and non-negative that is 0 ≤ P (A) ≤ 1 2. P(S) = 1. P(Ф) = 0 3. If A1, A2… An are sequence of any finite or infinite sequence of disjoint events of “S”, P (UAi) = Σ P (Ai) P(A1UA2UA3…UAn) = P(A1) + P(A2)… + P(An) Specifically for A1 and A2 alone P(A1UA2) = P(A1) + P(A2). for A1, A2 & A3 P(A1UA2UA3) = P(A1) + P(A2) + P(A3) [Image of overlapping circles] Disjoint events - P(A1UA2) = P(A1) + P(A2) If A1 and A2 are joint events P(A1UA2) = P(A1) + P(A2) - P(A1nA2) For A1, A2, A3 P(A1UA2UA3) = P(A1) + P(A2) + P(A3) - P(A1nA2) - P(A2nA3) - P(A1nA3) + P(A1nA2nA3) [Image of overlapping circles] In mathematical notation, probability of an event A is given by P(A) = n(A)/ n(S) ## Example If a fair coin is tossed 3 times, the sample space will be {HHH, HHT, HTH, THH, HIT, THI, ITH, TII}, what is the probability of obtaining even number of heads at the second trial? Soln: If a fair coin is tossed 3 times, the sample space will be S = {HHH, HHT, HTH, THH, HIT, THI, ITH, TII}, 2^3 = 8 Let A1 be an event of obtaining even number of heads. A2 be an event of H at 2nd trial A1 = {HHT, HTTH, THHI} n (A1) = 3 A2 = {HHH, HHT, THH, TTH} n (A2) = 4 P (A1 or A2) = P (A1UA2) Since A1 and A2 are joint events P(A1UA2) = P(A1) + P(A2) - P(A1nA2) now, A1UA2 = {HHT, HTTH, THHI, HHH}, P (A1nA2) = 2. ∴ P(A1UA2) = n(A1) + n(A2) - n (A1nA2) = 3 + 4 - 2 = 8 = n(S) = n(A1UA2)/ n(S) = 3 + 4 - 2/ 8 = 5/8 Therefore, if a fair coin is tossed 3 times, the probability of obtaining even number of heads at the 2nd trial is 5/8 ## Example 2 A bag contains 20 tickets marked with numbers 1 to 20. If a ticket is drawn at random, find the probability with be multiply: 1. 2 or 5 2. 3 or s Soln: Let the tickets be present by 12. S = {1, 2 ....20}, n (S) = 20 Let A1 be multiple of 2 A2 be multiple of 3 A3 be multiple of 5 A1 = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}, n (A1) = 10 A2 = {3, 6, 9, 12, 15, 18}, n (A2) = 6 A3 = {5, 10, 15, 20}, n (A3) = 4 ## Multiplication Theorem The ability of simultaneous happening of 2 out say Ak is given by P(A1nB1) = P (A1|B1)P (B1) where P (A1|B1) is the condition prob. of happening of A1 under the ## Independent Event The events say A & B are said to be independent of each other If the happening of any one of them is not affected by and does not affect the happening of any other event. If, A & B are two independent events P(A, B) = P(A) P(B) Let consider 3 events A, B and C If P(A, B) = P(A)P(B) P(A, C) = P(A)P(C) P(B, C) = P(B)P(C) and P (A, B, C) = P(A)P(B)P(C), then A, B and C are pairwise independent and mutually dependent event Let A & B outcome of an experiment and suppose that P(A) = 0.4, P(AUB) = 0.7 and P (B) = P. 1. For what choice of P and A & B are mutually exclusive? 2. For what choice of P are A & B independent? Hence, P(A) = 0.4, P(AUB) = 0.7, P(B) = P A and B can be mutually exclusive if ⇒ P(AUB) = P(A) + P(B) - P(A, B) Since A & B are mutually exclusive P(A, B) = 0 ⇒ 0.7 = 0.4 + P ∴ P = 0.3 If the value of P is 0.3 then A & B are mutually exclusive. If A and B are independent P (A, B) = P(A) P(B) P (A, B) = 0.4P From ©, 0.7 = 0.4 + P - (A, B) P(A, B) = 0.3P P(A, B) = 1 - 0.3 From © and *** above 0.4 = P - (1-0.3) 0.3 = P - 0.7 P = 0.3 + 0.7 P = 1 ## Example 2 The probability that a contractor gets a plumbing contract is 2/5 and the probability he will not get an electric contract is 9/10. If the probability of getting at least one contract is 7/10 what is the probability of getting both a plumbing contract & an electric contract if he has got an electric contract? Soln: Let E represent and event of getting plumbing contract. Ez represent E and event of getting electric contract. P(E) = 2/5, P(E2) = 1 - 9/10 = 1/10 P(EUE2) = P(E) + P(E2) - P(EnE2) 7/10 = 2/5 + 1/10 - P(EnE2) P(EnE2) = 2/5 + 1/10 - 7/10 = 1/15 ## Assignment The probability that a contractor gets a plumbing contract is 2/3 and the probability that he will not get an electric contract is 9/10. What is the probability of getting at least one contract. What is the probability of getting both a plumbing and an electric contract if he has got an electric contract?