Probability Examples and Laws for Mutually Exclusive and Independent Events

Example 4: Determine the probability of an odd number appears in a single toss of a fair die? Solution: There are 6 equally likely outcomes 1, 2, 3, 4, 5, 6. There are 3 cases where die comes up 1, 3, and 5. P(E) =? ?=? ? Example 5: Determine the probability of the sum 7 appears in a single toss of a pair dice? Solution: Each of the six faces of one die can be associated with each of the six faces of the other dice, so that there are 6.6 = 36 equally likely outcomes. There are 6 ways or cases of obtaining the sum 7 (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) P (E) = 6/36 = 1/6

Laws of probability: 1. For mutually exclusive events Two or more events are called mutually exclusive if the occurrence of any one of them excludes the occurrence of the others. Thus if ??and ??are mutually exclusive events,P (??.??) = 0. Note: If either ??or ??or both occur then P(??+ ??) =P(??) + ? ?? P(??.??) Not mutually exclusive events P(??+ ??) =P(??) + ? ?? mutually exclusive events Example 6: A box contain 4 black balls, 5 white, a ball is drawn randomly what is the probability that it will be black or white? Solution: P (B+W) = P (B) +P (W) =4 9+5 9=9 9= 1

Example 7: By tossing a dice what is the probability of getting an even number? Solution: P (2+4+6) = ? ?+? ?+? ? ?=? ?= ? Example 8: In a college 25% of student failed at Mathematics. 15% of student failed at chemistry. 10% failed in both (Math. + Chemistry) If we choose a student randomly, what is the probability that he failed at (Math. + Chemistry). Solution: P (? + ?) =P (?) + ? ? P (? .?) = o.25 + o.15 0.10 = 0.30

2. For independent events the probability is :- P (??.??) = P (??).? ?? 3. For dependent events the probability is :- P (??.??) = P (??). P (??/??) P (??.?? ??) = P (??). P (??/??) P (??/????) . P (??/???? ?? ?).

Example 9: Two boxes, the first one contains (4) white balls and (2) black. The second box contains (3) white balls and (5) black. A ball is drawn from each. What is the probability that the two balls are black? 1 ball drawn 1 ball drawn Solution: P (??. ??)= P (??). P (??) = ( 2 6 ).( 5 8) = 10 48

Example 10: A box contains (5) red balls, (3) black, if two balls drawn together [or respectively without replacement to the first ball in box] what is the probability that the two balls were black? Two balls drawn together Solution: P (??) =(3 8 ) P (??/??) = (2 7 ) P (?? .??) = P (??). P (??/??)= (3 8 ). (2 6 56 ) 7 )= (

Example 11: A park of vehicles contains of 25 Taxi cars and 10 privet if a 3 chooses randomly what the probability of being Taxi is? Solution: P (?? .?? .??) = P (??). P (??/??). P (??/????). = ( ?? ?? )(?? ?? )(?? ?? ). Example 12:A box contains 20 white ball and 10 red, a ball is drawn with replacement, then another ball is drawn what is the probability that the two balls were red? Solution: P (??) =Prob. that the 1st drawn ball is red. P (??) =Prob. that the 2nd drawn ball is red. P (????)= P (??). P (??) = (10 30 )(10 30 )= (1 9 )

Factorial: If n is positive integer value, then Factorial n, denoted by n! And defined as: n! = n. (n-1). (n-2) 1 Thus 5! = 5.4.3.2.1 = 120 4! 3! = (4.3.2.1). (3.2.1) = 144 Note: It is convenient to define:- 0! = 1! =1 n! = n. (n-1)! n ! = n. (n-1). (n -2)! Permutation: Suppose that we are given (n) distinct objects and wish to arrange r of these objects in a line. Since there are (n) ways of choosing the 1st object, and after, this is done, n-1 ways of choosing the 2ndobjects and finally n- r +1 ways of choosing the ?? objects, it follows by fundamental principle of counting that the number of different arrangements or permutations as they are often called, denoted by P (n, r), ??,? or ??? , is given by: ??? = n. (n-1). (n -2) ( n- r +1) = ?! (? ?)! We call ??? the number of permutations of n objects taken r at a time. If r = n, then ??? = n! Which is called n factorial.

Example 13: The numbers of different arrangements or permutations consisting of 2 letters each that can be formed from the 3 letters A, B and C? Solution: (? ?)! = ?! ?! ??? = ?! = 3.2=6 AB, BA, AC, CA, BC, CB. ?! ?! (?)! = 3.2 = 6 (? ?)! = If taken 3 letters at a time, the arrangement is ???= ABC, ACB, BAC, BCA, CAB, CBA .

When the number of permutations of n objects consist of groups of which?? are alike, ?? are alike, here of course, n = ??+??+ +?? . Thenthe number of different permutations of the objects is n???,??, ,??= ?! (??,??, ,??)! . Example 14:The numbers of different permutations of the 11 letters of the word MISSISSIPPI, which consist of 1M, 4I S, 4S S, and 2P S, is Solution: ??! ?!?!?!?! = 34654

Example 14: Five red, two white and three blue marbles are arranged in a row .If all the marbles of the same color are not distinguishable from each other, how many different arrangements are possible? Solution: ?? = 5 n = 10 ?? = 2 ?? = 3 ??! ?!?!?! = ??.?.?.?.?.?! = ??.?.?.?.? ?!?! Number of arrangements = = 2520 ?!?!?! Example 15: In how many ways can 10 vehicles be arrangement on a line if only 4 parks are a valuable? Solution: n = 10 r = 4 Number of arrangements = ???= ????= ??.?.?.?.?! = 5040 ?!

Example 16: Four different math. Books, Six Physics different books and two different chemistry books are to be arranged on shelf, how many different arrangement are possible if: 1-The books in each particular subject must all stand together. 2- Only the mathematics books must stand together. Solution: For the math. Books ???= 4! ways For the physics books: ???= 6! Ways For the chemistry books:???= 2! Ways For the three groups: ???= 3! Ways Number of arrangement = ???.???.???.???=4! 6! 2! 3! = 207360 For the math. Books ???= 4! ways Consider the four math. Books as one big book, then we have 9 books [1 math. + 6physics + 2 chemistry] = ??? = 9! Ways Number of arrangements = ??? .??? = 4! 9! = 8709120