Understanding Basics of Binomial Distribution

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Understanding Basics of Binomial Distribution :

Understanding Basics of Binomial Distribution There are situations in which we classify the parameter in to two classes: good ,bad/ yes,no /above a value ,below a value. Eg.product we buy is good or bad .Probability of being good is p=1/2. It means we take a large number (?) half of them are good .It does not mean if two are taken 1 is good. If we take say 100 of them what is the probability that 20 are good or given probability p how many to be picked that the probability all are bad is zero. These probiabilities Follow Binomial distribution . Common examples for explaining discrete probability distributions are ; 1.Tossing of a coin with 2 possible outcomes (2 faces);Head or Tail 2.Throwing of a dice with 6 possible outcomes (6 faces) The probability of a Head for a fair(unbiased) coin is ½.That means if we toss the Coin large number of times ,heads will show up half the number of times. What is the probability of getting a head and tail for a coin tossed in two successive Attempts.?

Probability of a Head and Tail:

Probability of a Head and Tail The possible outcomes are HH,TT,HT,TH,TT It is intutive to see P(HH)=P( tt )=.25,P(ht/ th )=.5 For a fair coin. What if the coin is biased for tail with P(t)=2/3and P(h)=1/3. Given this how do we compute P(ht/ th ),P( hh ) etc.

Computation of P(th/ht):

Computation of P( th /ht) P(h)=1/3 means if we toss 300 times ,heads will turnup 100 times. We toss the coin first time 300 times.For each of this we toss the coin 300 times second time ,resulting in a total of 300x300 tosses/trials. This gives the following arrangment of heads and tails in two successive tosses.

Distribution of TH,HT,HH,TT:

Distribution of TH,HT,HH,TT First Toss Second toss H 100x100 H H H - T T T T P(HH)=100x100/300x300=1/3X1/3=1/9 P(TT) =200x200/300x300=2/3x2/3=4/9 P(TH/HT)=100x200/300x300+200x100/300x300 =1/3x2/3+2/3x1/3=4/9 200x100 200x200 100x200

Binomial Distribution Generalized:

Binomial Distribution Generalized If P(H)=p P (T)=q=1-p because p+q =1 P(HH)=P(H) xP (H)= pxp P(TT)= qxq P(ht)= pxq P( th )= qxp ,P(ht/ th )=2pq With these preliminaries ,we proceed to obtain The general formulation to answer – What is the probability of occurrence “r” times one outcome out of the two possible outcomes in “n” successive attempts/trials?

Binomial distribution for “r” successes in “n” trials:

Binomial distribution for “r” successes in “n” trials Take the example of a coin ,we have the following possibilities in “n” trials pattern number probability ttttttt --- tt 1 q**n ttttttt --- th n pq **(n-1) ttttttt thh n(n-1)/2 p**2q**(n-2) ttttttt hhh n(n-1)(n-2)/3 p**3q**(n-3) n(n-1)(n-2)/3= P(H r times)= ncr p**R xq **(n-r)

Alternative derivation of Binomial distribution:

Alternative derivation of Binomial distribution p be the probability of head,q that of tail. A toss would result with probability 1 either head or tail 1= p+q;result of two tosses with probability 1 Two tosses 1=( p+q )( p+q ) Three tosses 1=( p+q )( p+q )( p+q ) n tosses 1=( p+q )**n Binomial expansion gives the probability of r heads In n trials as nCrp **r q**(n-r)

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