#total probability

[jacob-cs]

MIT RES.6-012 Introduction to Probability 10강

L10.2 Conditional PDFs

https://youtu.be/Kj6iEzXsFkI

L10.3 Comments on Conditional PDFs

https://youtu.be/mKcWk_DmS7M

L10.4 Total Probability & Total Expectation Theorems

https://youtu.be/0cD-tcITuck

L10.5 Independence

https://youtu.be/JCQnsPggTp8

L10.8 Bayes Rule Variations

https://youtu.be/WSrVCCBOeg4

L10.9 Mixed Bayes Rule

https://youtu.be/363JQxFwLXg

[workdevelopingpurchasep]

Application of Binomial Distribution

The aggregate in reference to successes for discrete probability distribution in sequence of ‚¬n' well provided for event is called for binomial distribution. The applications of trinomial distribution are worn as representing probability distribution on route to calculate using Bernoulli's formula. Binomial distribution is experimented in Bernoulli's theorem. Let the possibility in regard to getting a success in a single ordeal be there p and that of getting a poor stewardship hold q so that p + q = 1. <\p>

p(X =r) = ncr pr q(n-r) where p + q=1 then q =1-p<\p>

Engrossment all for Bicameral Distribution:<\p>

Docimasy as representing application of binomial distribution:<\p>

Proof:<\p>

Let us denote a success by s and a failure in agreement with F.<\p>

Kidney in connection with ways of getting r successes in n trials = ncr<\p>

P(X=r) = ncr.p} sss!.s and fff!f }<\p>

= ncr}p(s).p(s)!r r times} * }P(F)!P(F)!.(n-r)times}<\p>

= ncr(p.p.p!. r the present day) * (q.q.q!. (n-r) times)<\p>

=ncr pr q(n-r)<\p>

For that cause, P(X=r) = ncr pr q(n-r)<\p>

We have P(CROSS FOURCHEE=0) = qn; P(X = 1) = npq(n-1); P(X=2) = nc2.p2 q(n-2), etc.<\p>

The bias distribution of x may be expressed as<\p>

( X: 0 1 !!!.. r )<\p>

(P(X): qn npq(n-1)!!!!. Cr pr q(n-r))<\p>

This distribution is called binomial distribution.<\p>

Conditions on account of the applicability respecting a binomial distribution<\p>

i) The experiment is performed as things go a strait and fixed number of trials.<\p>

ii) Several trial must give either a easy street or a failure.<\p>

iii) The probability of a success swank each trial is the same.<\p>

Total probability = nCr pr qn-r<\p>

= nC1 p0 qn + nC1 p1 qn-1 +.......................+ nCn pn q0<\p>

= ( q + p )n<\p>

= ( 1 )n<\p>

= 1<\p>

Examples in furtherance of Application of Binomial Disrtibution:<\p>

For example 1:<\p>

A die is tossed 4 times. What is the Probability of getting exactly 3 fours?<\p>

Solution:<\p>

Hereunto n = 4, x = 3, odds of success on a single trial = 1\ 4 or 0.25.<\p>

Therefore, The binomial probability is,<\p>

p(CRUX DECUSSATA =r) = ncr pr q(n-r) where p + q=1 old q =1-p<\p>

q=1-0.25<\p>

q=0.75<\p>

b( 3; 4, 0.25 ) = 4C3 €" ( 0.25)3 €" ( 0.75)4 †'3<\p>

= ( 4! \ 3! €" (4-3)!) €" 0.016 €" ( 0.75)<\p>

= (4! \ 3! €" 1!) €" 0.016€" 0.75<\p>

= 4 €" 0.016 €" 0.75<\p>

b( 3; 4, 0.25 ) = 0.048. Answer.<\p>

Example 2:<\p>

A die is tossed 7 times. What is the Likelihood of getting exactly 5 twos?<\p>

Solution:<\p>

Here n = 7, x = 5, happenstance of star on a simple probative = 1\ 7 or 0.143.<\p>

Therefore, The binomial statistical prediction is,<\p>

p(X =r) = ncr pr q(n-r) where p + q=1 then q =1-p<\p>

q=1-0.143<\p>

q=0.857<\p>

b( 5; 7, 0.143 ) = 7C5 €" ( 0.143)5 €" ( 0.857)7 †'5<\p>

= ( 7! \ 5! €" (7-5)!) €" (0.0001) €" ( 0.857)2<\p>

= (7! \ 5! €" 2!) €" 0.0001€" 0.734<\p>

= 21 €"0.0001 €" 0.734<\p>

b( 5; 7, 0.143 ) = 0.00154. Answer.<\p>

Practice Problems for Application of Bicuspid Distribution:<\p>

Practice problems:<\p>

1) A die is tossed 6 this point. What is the Probability of getting exactly 2 fours?<\p>

2) A die is tossed 2 times. What is the Probability in respect to getting rigorously 1sixes?<\p>

Answers:<\p>

1) 0.201<\p>

2) 0.5.<\p>