Topic--Probability CBSE CLASS 12 : Topic--Probability CBSE CLASS 12 1 copyrighted by abhijit banerjee
Topic 1Conditional ProbabilityFor any two events A and B with P(B) ? 0 , the conditional probability of A given that B has occurred is defined by: P(A|B) = P(A ? B)/P(B)PROPERTIES1.Let F be the only event of a sample space S of an experiment, then we have P (S|F) =P(F|F) = 1 : Topic 1Conditional ProbabilityFor any two events A and B with P(B) ? 0 , the conditional probability of A given that B has occurred is defined by: P(A|B) = P(A ? B)/P(B)PROPERTIES1.Let F be the only event of a sample space S of an experiment, then we have P (S|F) =P(F|F) = 1 2 copyrighted by abhijit banerjee
Slide 3 : 3 copyrighted by abhijit banerjee
Slide 4 : 4 copyrighted by abhijit banerjee
2.If A and B are any two events of a sample space S and F is an event of S such that P(F) ? 0, then P((A ? B)|F) = P(A|F) + P(B|F) – P ((A n B)|F) In particular, if A and B are disjoint events, then P((A?B)|F) = P(A|F) + P(B|F) : 2.If A and B are any two events of a sample space S and F is an event of S such that P(F) ? 0, then P((A ? B)|F) = P(A|F) + P(B|F) – P ((A n B)|F) In particular, if A and B are disjoint events, then P((A?B)|F) = P(A|F) + P(B|F) 5 copyrighted by abhijit banerjee
Slide 6 : 3.P(E'|F) = 1 - P (E|F) Examples:-
1.If P (A) =7/13 , P (B) =9/13 and
P (A n B) = 4/13 ;evaluate P(A|B)
Ans:-P(A|B)
=P (A n B) |P(B)
={(4/13)|(9/13)}=4/9 6 copyrighted by abhijit banerjee
Slide 7 : 3.P(E'|F) = 1 - P (E|F) Examples:-
2. A family has two children. What is the probability that both the children are boys given that at least one of them is a boy ?
Ans:- sample spaces =4 like
S = {(b, b), (g, b), (b, g), (g, g)} [b=boy; g=girl]
E = {(b,b)} [event of both are boys]=1
F = {(b,b), (g,b), (b,g)}[ any of them is boy] =3
E n F = {(b,b)} ;
hence
P(E|F)= P(E n F) | P(F)
={(1/4) | (3/4)}= 1/3 7 copyrighted by abhijit banerjee
Slide 8 : Examples for practice:-
A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?
2.Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that(i) the youngest is a girl, (ii) at least one is a girl? 8 copyrighted by abhijit banerjee
Topic 2:-Multiplication Theorem WE KNOWP(E|F) =P(E n F) | P(F) [ when P(F) ? 0 ]hence;P(E n F) = P(F) . P(E|F)This above formula is known as multiplication rule in probability. : Topic 2:-Multiplication Theorem WE KNOWP(E|F) =P(E n F) | P(F) [ when P(F) ? 0 ]hence;P(E n F) = P(F) . P(E|F)This above formula is known as multiplication rule in probability. 9 copyrighted by abhijit banerjee
Topic 2:-Multiplication TheoremExamples1.An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?Ans- : Topic 2:-Multiplication TheoremExamples1.An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?Ans- P(E) = P (black ball in first draw) = 10/15 Also given that the first ball drawn is black, i.e., event E has occurred, now there are 9 black balls and five white balls left in the urn.
P(F|E)=9/14
Now from multiplication rule
P(E n F) = P(E) . P(F|E)=(10/15) ×(9/14)=3/7 Let E and F denote respectively the events that first and second ball drawn 10 copyrighted by abhijit banerjee
Topic 2:-Multiplication TheoremExamples1.An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?Ans- : Topic 2:-Multiplication TheoremExamples1.An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?Ans- P(E) = P (black ball in first draw) = 10/15 Also given that the first ball drawn is black, i.e., event E has occurred, now there are 9 black balls and five white balls left in the urn.
P(F|E)=9/14
Now from multiplication rule
P(E n F) = P(E) . P(F|E)=(10/15) ×(9/14)=3/7 Let E and F denote respectively the events that first and second ball drawn 11 copyrighted by abhijit banerjee
Topic 2:-Multiplication TheoremExamples : Topic 2:-Multiplication TheoremExamples Multiplication rule of probability for more than two events If E, F and G are three events of sample space, we have P(E n F n G)= P (E) P (F|E) P(G|(E n F))= P (E) P(F|E) P(G|EF)
Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards. What is the probability that first two cards are kings and the third card drawn is an ace?
Ans:-
we have to find P (KKA)=P(K) P(K|K) P(A|KK)
=(4/52) (3/51) (4/50) =2/5525
[ P(K)= probability of first card is king among 52 cards;
P(k/k)= probability that the 2nd card king among 51 cards;
P(A/KK)=probability when the 3rd card is aces among 50 cards] 12 copyrighted by abhijit banerjee
Conclusion If u like my slides and want coaching in math subjects please contact me at My mail addresswishabhijit_ju@rediffmail.com.Online Coaching is available through wiziq and through skype.I will help you in solving difficult problems given in your books through online. CBSE,ICSE,ISC 9-12TH Grade any problems.U can add me in your contact through wiziq.I will take some public classes shortly in this month.So hurry contact now for free classes. I have great collection of problems and all materials like this are prepared by me for hours to give u benefit.For my profile u can visit my wiziq profile. : Conclusion If u like my slides and want coaching in math subjects please contact me at My mail addresswishabhijit_ju@rediffmail.com.Online Coaching is available through wiziq and through skype.I will help you in solving difficult problems given in your books through online. CBSE,ICSE,ISC 9-12TH Grade any problems.U can add me in your contact through wiziq.I will take some public classes shortly in this month.So hurry contact now for free classes. I have great collection of problems and all materials like this are prepared by me for hours to give u benefit.For my profile u can visit my wiziq profile. 13 copyrighted by abhijit banerjee