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Modern Math
PROBABLITY  2
CONDITIONAL PROBABLITY & BAYES THEOREM
If occurrence of one event out of two or more events from the sample space affects the occurrence of other, this is called conditional probability.
Let A and B be two events associated with an experiment. Then the probability of occurrence of event A if B has already occurred and is called conditional probability. It is denoted by P(A/B).
The elements of B which favor the event A are the common elements of A and B. i.e, the sample points of .
It is valid only when
Properties:
Let E and F events of sample space S of an experiment, then
1.
If A and B are any two events of a sample space S and F in an event of S such that P(F) 0, then
2.
If E and E’ are disjoint events
3.
INDEPENDENT EVENTS:
If occurrence of one event does not affect the occurrence of another events are called independent events. In other words, Two events are said to be independent, if
P(A/B) = P(A) provided P(B) 0
P(B/A) = P(B) provided P(A) 0
By multiplication rule of probability, If A and B are two independent events, then
Bayes Theorem:
If E1, E2 , ……En are n non empty events which constitute a partition of the sample space S i.e. E1 , E2 , E3…. En are pairwise disjoint and E1E2E3...... En = S and A is any event of non zero probability, then the probability of occurrence of Ei while A is already occurred is
Example 1: A die is rolled. If the outcome is an odd number, what is the probability that it is a prime?
Solution:
When a die is rolled, the outcomes are S= {1, 2, 3, 4, 5, 6}
Let A = event of getting an odd number
B = event of getting a prime number
Then, A = {1, 3, 5}
B= {2, 3, 5}
= {3, 5}
P(A) = 3/6 = 1/2
P (B) = 3/6 = 1/2
P= 2/6 = 1/3
Required Probability
Example 2: In a class, 30% students study English, 15% study Hindi and 10% study both Hindi and English. One student is selected at random. Find the probability that he studies English if it is known that he studies Hindi.
Solution:
E = event of studying English and
H = Event of studying Hindi
P(E) = 30/100
P(H) = 15/100
P(EH)= 10/100
Required probability =
Example 3: Two numbers are selected random from integers 1 to 9. If the sum of numbers selected is even, find the probability that both the numbers are odd.
Solution:
A= Event of choosing two odd numbers
B = Event of choosing two numbers whose sum is even
The number of ways in which 2 odd numbers can be selected out of 5 = 5C2
The number of ways of choosing 2 odd numbers out of 1 to 9 whose sum is even = 4C2 + 5C2
P (B) =
P (AB) =
The required probability,
Example 4: A box contains 20 red and 15 black balls. Two balls are drawn from the box one after the other without replacement. What is the probability that both drawn balls are red?
Solution:
The number of ways in which two balls can be drawn without replacement = 35C1 * 34C1
The number of ways in which two red balls can be drawn = 20C1 * 19C1
The required probability =
Example 5: A bag contains 4 white and 2 black balls . Another bag contains 3 white and 5 black balls. If one ball is drawn from each bag, find the probability that

Both are white

One is white, one is black.
Solution:
Bag 1 
Bag 2 
White 4 
White 3 
Black 2 
Black 5 
P(W1) = 4/6 = 2/3 
P(W2) = 3/8 
P(B1) = 2/6 = 1/3 
P(B2) = 5/8 

Required probability =

Required probability =
Example 6: A man is known to speak the truth 4 out of 5 times. He throws a die and reports that it is a four. Find the probability that it is actually a four.
Solution:
In a throw of die, suppose
E1= Event of getting a four
E2= Event of getting not four
E = event that the man reports that it is four
P (E1) = 1/6
P (E2) = 11/6 = 5/6
P(E/E1) = probability that man speaks truth = 4/5
P(E/E2) = probability that man does not speak truth = 1/5
Required probability= probability of getting four, given that man reports it to be a four
Example 7: Bag ‘A’ contains 9 cards numbered 1 through 9, and bag B contains 5 cards numbered 1 through 5. A bag is selected random and a card is drawn from the bag. If the number is even, find the probability that card come from bag A
Solution:
Probability of selecting a bag randomly P(A)= 1/2
Probability of selecting an even number from bag A i.e. P(E/A)= 4/9
Probability of selecting an odd number from bag A i.e. P(O/A)= 5/9
Probability of selecting an even number from bag B i.e. P(E/ B)= 2/5
Probability of selecting an odd number from bag B i.e. P(O/B)= 3/5
We have to find P(A/E) = probability that bag ‘A’ was selected, if it is known that the number is even.
Example 8: A person goes to his office by using different means of transport on different days. It is known that probabilities that he will come by train, bus scooter or by car are respectively 1/10, 3/10, 2/10 and 4/10. The probability that he will be late are ¼, 1/5, 1/6 and 1/3 if he comes by train , bus , scooter and car respectively, on one day when he reaches office, he is late. What is probability he come by train?
Solution:
Suppose E1, E2, E3 and E4 be events that person come by train , bus , scooter and care respectively.
Given that,
P(E1) = 1/10
P(E2) = 3/10
P(E3) = 2/10
P (E4) = 4/10
Let E is the event that person is late. Then,
P(E/E1) = 1/4
P(E/E2) = 1/5
P(E/E3) = 1/6
P(E/E4) = 1/3
Probability that he comes by train, given that he is late
Example 9: A bag contains a fair coin ‘A’ and a two headed coin’ B’. A coin is selected at random and tossed twice. If it heads appears both times find the probability that coin is two headed.
Solution:
Probability of selecting a coin random = 1/2
If Coin ‘A’ is selected, then
Probability of getting head both times (HH) = 1/4
Probability of getting (HT) = 1/4
Probability of getting (TT) = 1/4
Probability of getting (TH) = 1/4
If Coin ‘B’ is selected, then
Probability of getting heads = 1
Required probability =
Example 10: A factory has three machines A, B and C. Producing 100, 200 and 500 bolts per day respectively. The machine A produces 1% defective bolts, B produces 2% defective bolts and C produces 5 % defective bolts. At the end of day, a bolt is drawn random and it is found defective. What is the probability that this defective bolt has been produced by machine ‘A’ ?
Solution:
Total number of bolts produced in a day = 100+200+500 = 800
Suppose E1, E2 and E3 be the events of drawing a bolt produced by machines ‘A’ , ‘B’ and ‘C’ respectively, then
P(E1) = 100/800 = 1/8
P(E2) = 200/800 = 2/8
P (E3) = 500/800 = 5/8
Let E be the event of drawing a defective bolt. Then,
P(E/E1) = 1/100
P(E/E2) = 2/100
P(E/E3) = 5/100
Required probability