Binomial Distribution

Transcription

1 Binomial Distribution

2 Probability Eeriment: tossing an unfair coin two times. Probability of success = Probability of getting a tail = 0.7 Probability of failure = Probability of getting a head = 0.3 Number of Trials (n) = 2 Total number of ossible outcomes = 2 2 = 4 Samle Sace of Possible Outcomes = {TT, TH, HT, HH} = {SS, SF, FS, FF} Samle Sace Successes (S) Failures (F) Probability of Each Outcome SS 2 0 (0.7)(0.7) = 0.49 SF 1 1 (0.3)(0.7) = 0.21 FS 1 1 (0.7)(0.3) = 0.21 FF 0 2 (0.3)(0.3) = 0.09 What is the robability of obtaining 0 success? P(FF) = P(HH) = P(H) P(H) = 0.09 What is the robability of obtaining 1 success? Outcomes with one success = {SF, FS} Probability of obtaining 1 success = P(SF) + P(FS) = = 0.42 What is the robability of obtaining 2 successes? Outcomes with two successes = {SS} Probability of obtaining two successes = P(SS) = 0.49

3 Binomial Distribution Probability Eeriment: tossing an unfair coin three times. Probability of success = Probability of getting a tail = 0.6 Probability of failure = Probability of getting a head = 0.4 Number of Trials (n) = 3 Total number of ossible outcomes = 2 3 = 8 Samle Sace of Possible Outcomes = {SSS SSF SFS SFF FSS FSF FFS FFF} Samle Sace Successes (S) Failures (F) Probability of Each Outcome SSS 3 0 (0.6)(0.6)(0.6) = SSF 2 1 (0.6)(0.6)(0.4) = SFS 2 1 (0.6)(0.4)(0.6) = SFF 1 2 (0.6)(0.4)(0.4) = FSS 2 1 (0.4)(0.6)(0.6) = FSF 1 2 (0.4)(0.6)(0.4) = FFS 1 2 (0.4)(0.4)(0.6) = FFF 0 3 (0.4)(0.4)(0.4) = What is the robability of obtaining 0 success? P(FFF) = P(F) P(F) P(F) = = 0.064

4 What is the robability of obtaining 1 success? Outcomes with one success = {SFF FSF FFS} Number of outcomes with one success = 3 Probability of obtaining 1 success = P(SFF) + P(FSF) + P(FFS) = = Alternative Method: Binomial Formula: P( )! n! 1 n 3 number of trials 1 number of success(es) 0.6 robability of success n P( 1) 1! n! 3! 1! 3 1! n

5 What is the robability of obtaining 2 successes? Outcomes with two successes = {SSF SFS FSS } Number of outcomes with two successes = 3 Probability of obtaining 2 successes = P(SSF) + P(SFS) + P(FSS) = = Alternative Method: Binomial Formula: P( )! n! 1 n 3 number of trials 2 number of success(es) 0.6 robability of success n P( 2) 1! n! 3! ! 3 2! n What is the robability of obtaining 3 successes? Outcome(s) with three successes = {SSS} Number of outcomes with three successes = 1

6 Probability of obtaining 3 successes = P(SSS) = Alternative Method: Binomial Formula: P( )! n! 1 n 3 number of trials 3 number of success(es) 0.6 robability of success n P( 3) 1! n! 3! ! 3 3! (1) n Probability Distribution Number of Success () P() P( )

7 Eected Value = P( ) 1.8 Eected Value = n (3)(0.6) 1.8 For Binomial Distribution: = Eected Value = n = Standard Deviation = n 1

8 Binomial Distribution Probability Eeriment: tossing an unfair coin four times. Probability of success = Probability of getting a tail = 0.2 Probability of failure = Probability of getting a head = 0.8 Number of Trials (n) = 4 Total number of ossible outcomes = 2 4 = 16 Samle Sace of Possible Outcomes = {SSSS SSSF SSFS SFSS FSSS SSFF SFSF FSSF FSFS FFSS SFFF FSFF FFSF FFFS SFFS FFFF} Samle Sace Successes (S) Failures (F) Probability of Each Outcome SSSS 4 0 (0.2)(0.2)(0.2)(0.2) = SSSF 3 1 (0.2)(0.2)(0.2)(0.8) = SSFS 3 1 (0.2)(0.2)(0.8)(0.2) = SFSS 3 1 (0.2)(0.8)(0.2)(0.2) = FSSS 3 1 (0.8)(0.2)(0.2)(0.2) = SSFF 2 2 (0.2)(0.2)(0.8)(0.8) = SFSF 2 2 (0.2)(0.8)(0.2)(0.8) = SFFS 2 2 (0.2)(0.8)(0.8)(0.2) = FSSF 2 2 (0.8)(0.2)(0.2)(0.8) = FSFS 2 2 (0.8)(0.2)(0.8)(0.2) = FFSS 2 2 (0.8)(0.8)(0.2)(0.2) = SFFF 1 3 (0.2)(0.8)(0.8)(0.8) = FSFF 1 3 (0.8)(0.2)(0.8)(0.8) = FFSF 1 3 (0.8)(0.8)(0.2)(0.8) = FFFS 1 3 (0.8)(0.8)(0.8)(0.2) = FFFF 0 4 (0.8)(0.8)(0.8)(0.8) =

9 What is the robability of obtaining 0 success? P(FFFF) = P(F) P(F) P(F) P(F) = = What is the robability of obtaining 1 success? Outcomes with one success = {SFFF FSFF FFSF FFFS} Number of outcomes with one success = 4 Probability of obtaining 1 success = P(SFFF) + P(FSFF) + P(FFSF) + P(FFFS) = = Alternative Method: Binomial Formula: P( )! n! 1 n 4 number of trials 1 number of success(es) 0.2 robability of success n P( 1) 1! n! 4! 1! 4 1! n

10 What is the robability of obtaining 2 successes? Outcomes with three successes = {SSFF SFSF SFFS FSSF FSFS FFSS} Number of outcomes with one success = 6 Probability of obtaining 2 successes = P(SSFF) + P(SFSF) + P(SFFS) + P(FSSF) + P(FSFS) + P(FFSS) = = Alternative Method: Binomial Formula: P( )! n! 1 n 4 number of trials 2 number of success(es) 0.2 robability of success n P( 2) 1! n! 4! ! 4 2! n What is the robability of obtaining 3 successes?

11 Outcomes with three successes = {SSSF SSFS SFSS FSSS } Number of outcomes with one success = 4 Probability of obtaining 3 successes = P(SSSF) + P(SSFS) + P(SFSS) + P(FSSS) = = Alternative Method: Binomial Formula: P( )! n! 1 n 4 number of trials 3 number of success(es) 0.2 robability of success n P( 3) 1! n! 4! ! 4 3! n

12 What is the robability of obtaining 4 successes? Outcomes with three successes = {SSSS} Number of outcomes with 4 success = 1 Probability of obtaining 4 successes = P(SSSS) = Alternative Method: Binomial Formula: P( )! n! 1 n 4 number of trials 4 number of success(es) 0.2 robability of success n P( 4) 1! n! 4! ! 4 4! (1) n

13 Probability Distribution Number of Success () P() P( ) Eected Value = P( ) = 0.8 Eected Value = n (4)(0.2) 0.8 For Binomial Distribution: = Eected Value = n = Standard Deviation = n

14 Binomial Distribution Probability Eeriment: tossing an unfair coin five times. Probability of success = Probability of getting a tail = 0.2 Probability of failure = Probability of getting a head = 0.8 Number of Trials (n) = 5 Total number of ossible outcomes = 2 5 = 32 Samle Sace Successes (S) Failures (F) Probability of Each Outcome SSSSS 5 0 (0.2)(0.2)(0.2)(0.2)(0.2) = SSSSF 4 1 (0.2)(0.2)(0.2)(0.2)(0.8) = SSSFS 4 1 (0.2)(0.2)(0.2)(0.8)(0.2) = SSFSS 4 1 (0.2)(0.2)(0.8)(0.2)(0.2) = SFSSS 4 1 (0.2)(0.8)(0.2)(0.2)(0.2) = FSSSS 4 1 (0.8)(0.2)(0.2)(0.2)(0.2) = SSSFF 3 2 (0.2)(0.2)(0.2)(0.8)(0.8) = SSFSF 3 2 (0.2)(0.2)(0.8)(0.2)(0.8) = SSFFS 3 2 (0.2)(0.2)(0.8)(0.8)(0.2) = SFSSF 3 2 (0.2)(0.8)(0.2)(0.2)(0.8) = SFSFS 3 2 (0.2)(0.8)(0.2)(0.8)(0.2) = SFFSS 3 2 (0.2)(0.8)(0.8)(0.2)(0.2) = FSSSF 3 2 (0.8)(0.2)(0.2)(0.2)(0.8) = FSSFS 3 2 (0.8)(0.2)(0.2)(0.8)(0.2) = FSFSS 3 2 (0.8)(0.2)(0.8)(0.2)(0.2) = FFSSS 3 2 (0.8)(0.8)(0.2)(0.2)(0.2) = SSFFF 2 3 (0.2)(0.2)(0.8)(0.8)(0.8) = SFSFF 2 3 (0.2)(0.8)(0.2)(0.8)(0.8) = SFFSF 2 3 (0.2)(0.8)(0.8)(0.2)(0.8) = SFFFS 2 3 (0.2)(0.8)(0.8)(0.8)(0.2) = FSSFF 2 3 (0.8)(0.2)(0.2)(0.8)(0.8) =

15 FSFSF 2 3 (0.8)(0.2)(0.8)(0.2)(0.8) = FSFFS 2 3 (0.8)(0.2)(0.8)(0.8)(0.2) = FFSSF 2 3 (0.8)(0.8)(0.2)(0.2)(0.8) = FFSFS 2 3 (0.8)(0.8)(0.2)(0.8)(0.2) = FFFSS 2 3 (0.8)(0.8)(0.8)(0.2)(0.2) = SFFFF 1 4 (0.2)(0.8)(0.8)(0.8)(0.8) = FSFFF 1 4 (0.8)(0.2)(0.8)(0.8)(0.8) = FFSFF 1 4 (0.8)(0.8)(0.2)(0.8)(0.8) = FFFSF 1 4 (0.8)(0.8)(0.8)(0.2)(0.8) = FFFFS 1 4 (0.8)(0.8)(0.8)(0.8)(0.2) = FFFFF 0 5 (0.8)(0.8)(0.8)(0.8)(0.8) = Find the robability of getting 0 success. n 5 number of trials 0 number of success(es) 0.2 robability of success P( 0) 1! n! 5! ! 5 0! n

16 Find the robability of getting 1 success. n 5 number of trials 1 number of success(es) 0.2 robability of success P( 1) 1! n! 5! 1! 5 1! n Find the robability of getting 2 successes. n 5 number of trials 2 number of success(es) 0.2 robability of success P( 2) 1! n! 5! ! 5 2! n

17 Find the robability of getting 3 successes. n 5 number of trials 3 number of success(es) 0.2 robability of success P( 3) 1! n! 5! ! 5 3! n Find the robability of getting 4 successes. n 5 number of trials 4 number of success(es) 0.2 robability of success P( 4) 1! n! 5! ! 5 4! n

18 Find the robability of getting 5 successes. n 5 number of trials 5 number of success(es) 0.2 robability of success P( 5) 1! n! 5! ! 5 0! n Probability Distribution Number of Success () P() P( ) Eected Value = P( ) 1.0 Eected Value = n (5)(0.2) 1.0

19 For Binomial Distribution: = Eected Value = n = Standard Deviation = n 1

20 Find the robability of getting at least 3 successes. " at least 3 successes" means 3 or more successes. P(at least 3 successes) = P(=3) + P(=4) + P(=5) = = Find the robability of getting at most 3 successes. " at most 3 successes" means less than or equal 3 successes. P(at most 3 successes) = P ( 3) = P(=0) + P(=1) + P(=2) + P(=3) = = Find the robability of getting at least 1 success. " at least 1 success" means 1 or more success(es). P(at least 1 success) = P ( 1) = P(=1) + P(=2) + P(=3) + P(=4) + P(=5) = =

21 Note that the comlement of "at least 1 success" is "no success". P(at least 1 success) = 1 - P(no success) = 1 - P(=0) = = According to the Pew Forum, the ercentage of eole living in the U.S. who identify themselves as rotestant is 51.3%. A random samle of 30 eole is selected, and were asked about their religious affiliation. a) Of the 30 eole randomly selected, how many are eected to identify themselves as rotestant? b) What is the robability that eactly 10 eole will identify themselves as rotestant? c) What is the robability that at least 10 eole will identify themselves as rotestant? d) What is the robability that at most 10 eole will identify themselves as rotestant? Solution: a) Eected Value = = n (30)(0.513) 15.39

22 b) n 30; 10 30! P(=10) = (1 ) r! n r! 10! 30 10! n c) P( 10) = P(=10) + P(=11) +P(=12) + P(=28) P(=29) P(=30) = d) P( 10) = P(=0) + P(=1) +P(=2) + P(=8) P(=9) P(=10) =

23 Eamle 2: Suose the unemloyment rate for those aged 16 to 19 in a northern state is 24 ercent. If a samle of 10 eole between the ages of 16 and 19 is selected at random. What is the robability that 8 of them are unemloyed? Probability that 8 of them are unemloyed 10! 8 2 = P(=8) = (1 ) ! n! 8! 10 2! n

24 Eamle 3: ZDNet reorted that 56% of emloyers check alicants rofiles on social networks like Facebook, LinkedIn, and Twitter. If 20 emloyers are selected at random, what is the robability that 12 of them check alicants rofiles on social networks? Probability that 12 of them check alicants' rofiles on social networks 20! 12 8 = P(=12) = (1 ) ! n! 12! 20 12! n

25 Eamle 4: The Economic Times reorted that 45% of residential rojects launched before 2008 in NCR (national caital region), Mumbai and Bangalore are delayed. If 40 residential rojects are selected at random, what is the robability that 10 of them are delayed? Probability that 10 of them are delayed 40! = P(=10) = (1 ) ! n! 10! 40 10! n