MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Menu
Get instant 20% OFF on Online Material.
coupon code: MOB20 | View Course list

Get extra R 440 off
USE CODE: MOB20

				   

Please explain the law of total probability?

7 years ago

Share

Answers : (1)

										

LAW OF TOTAL PROBABILITY:




The law of total probability is implied to the events that are mutually not exclusive. We can define this law as:

If n events A1, A2,……, An are n mutually exclusive events, then the probability of the happening of at least one of the events is the sum of the probabilities of the individual events. In symbols,

       


P [ A1+A2+A3+……+An ] = P[A1] + P[A2] + P[A3] + …… + P[An]


Or, 


P [ A1 U A2 U …… U An ] = P[A1] + P[A2] + P[A3] + …… + P[An]


Now, we discuss the law of total probability in details:


Let N be the total number of mutually exclusive, exhaustive and equally likely events out of which m1 cases are favourable to A1, m2 cases are favourable to A2 and so on.


The probability of occurrence of the event A1 = P[A1] = m1/N


The probability of occurrence of the event A2 = P[A2] = m2/N

Similarly, for A3, A4, A5, ………


The probability of occurrence of the event An = P[An] = mn/N



The events being mutually exclusive and equally likely, the total number of cases favourable to the event A1 or A2 or … or An is

m = m1 + m2 + m3 + …… + mn

Thus,

P [ A1+A2+A3+……+An] = m/N


= ( m1 + m2 + m3 +…… + mn ) / N


= (m1/N) + (m2/N) + …… + (mn/N)


= P[A1] + P[A2] + P[A3] + …… + P[An]


Hence, P [ A1 + A2 + …… + An ] = P[A1] + P[A2] + …… + P[An].


This is the law of total probability.


Alternatively, the law can be understood or interpreted as follows:  


If an event A consists of n mutually exclusive forms A1, A2, ……, An so that A happens whenever any one of the these events happens and conversely, then


A = A1+A2+A3+……+An and P[A] = P[A1] + P[A2] + P[A3] + …… + P[An].


The Law of Total Probability allows us to evaluate probabilities of events that are difficult to obtain alone, but become easy to calculate once we condition on the occurrence of a related event. Initially, we assume that the related event occurs, and subsequently, that it does not occur. The resulting conditional probabilities help us compute the total probability of occurrence of the event of interest.

The limitation of the law of total probability is that it cannot be used to evaluate the probabilities of the events that are not mutually exclusive.

7 years ago

Post Your Answer

Other Related Questions on Algebra

If alpha is a real root of the equation ax 2 +bx+c and beta is a real root of equation -ax 2 +bx+c. Show that there exists a root gama of the equation (a/2)x 2 +bx+c which lies between alpha...
 
 
 
Ajay 6 months ago
 
Small Mistake in last para posting again..............................................................................................................
 
Ajay 6 months ago
 
We have Similarly, So if P(x) = a/2 x 2 +bx +c, then and are off opposite sign and hence there must exist a root between the two numbers.
 
mycroft holmes 6 months ago
In the listed image can you tell me how beta*gamma = 2 ….. . . .. ??
 
 
The value of gamma is still not correct, either printing mistake or you gave me wrong value. The correct value of gamma is below
 
Ajay 5 months ago
 
Thankyou so much............................. …......................................................................!
 
Anshuman Mohanty 5 months ago
 
Yes sorry..... . . . .it is not so clear.. ok the values are beta = α + α^2 + α^4 and gamma = α^3 + α^5 + α^7
 
Anshuman Mohanty 5 months ago
if |z - i| Options: a*) 14 b) 2 c) 28 d) None of the above
 
 
If |z-i| = ?? PLs complete the question
  img
Nishant Vora one month ago
 
Got it! [z + 12 – 6 i ] can be rewritten as [ z – i + 12 – 5 i] => | z – i | and => |12 – 5 i | = sqrt ( 12^2 + 5^2) = 13......................(2) => | z + 12 – 6 i | => | z + 12 – 6 i |...
 
Divya one month ago
 
I tried posting the question several times, it kept cutting off the rest of the question. Here: If | z-1| Options: a*) 14 b) 2 c) 28 d) None of the above
 
Divya one month ago
sin^2 6°-sin^2 12°+sin^2 18°-sin^2 24°......15 solve it Urgent
 
 
Ajay, the complete qution isSolution is sin^2 6°-sin^2 12°+sin^2 18°-sin^2 24°..... upto 15 terms. sin 78°=0 sin 42°+sin 54°+ sin 66°+ + sin 18° sin 6°+ where )=0.5 (your required answer),...
 
Kumar 3 months ago
 
Not any people get my answer why. You can no give answer my question I am join this site
 
Vivek kumar 5 months ago
 
Hello If you want to get the solution quick you should post your question in clear manner. Its not clear what you wnat us to solve, and what does 15 at the end of question means?
 
Ajay 5 months ago
Solve: (sin theta+cosec theta)^2 + (cos theta +sec theta)^2- (tan^2 theta + cot ^2 theta)^2
 
 
What needs to be solved here ? The question is incomplete....................................................................
 
Ajay 6 months ago
 
i don’t know how to do this...............................................................................................
 
Saravanan 2 months ago
 
this is the question :: Solve: (sin theta+cosec theta)^2 + (cos theta +sec theta)^2 - (tan^2 theta + cot ^2 theta)
 
Naveen Shankar 6 months ago
solutions to Question no. 17,18 19 and 20 pleaseeeeeeeeeee
 
 
Let the feet of the altitudes on BC, AC, AB, be D,E,F resp. Let the orthocenter be H. The following can be proved easily: ​1. HDCE and HFBD are cyclic quadrilaterals. Then chord HE subtends...
 
mycroft holmes one month ago
 
Draw which is Isoceles as OB = OC. Now which means . Let D, be the foot of the perp from O on BC ( which is also the midpoint of BC). Then OD = OC sin (OBC) = R cos A. Hence the required...
 
mycroft holmes one month ago
 
a cos A = b cos B 2R sin A cos A = 2R sin B cos B sin 2A = sin 2B Either A = B (isoceles or equilateral) or 2A = 180 o – 2B so that A+B = 90 o .(Right-angled)
 
mycroft holmes one month ago
View all Questions »

  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: R 15,000
  • View Details
Get extra R 3,000 off
USE CODE: MOB20

Get extra R 440 off
USE CODE: MOB20

More Questions On Algebra

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!!
Click Here for details