<< Chapter < Page Chapter >> Page >
P ( A B C ) = P ( A ) P ( A B ) P ( A ) P ( A B C ) P ( A B ) = P ( A ) P ( B | A ) P ( C | A B )

and

P ( A B C D ) = P ( A ) P ( A B ) P ( A ) P ( A B C ) P ( A B ) P ( A B C D ) P ( A B C ) = P ( A ) P ( B | A ) P ( C | A B ) P ( D | A B C )

This pattern may be extended to the intersection of any finite number of events. Also, the events may be taken in any order.

Selection of items from a lot

An electronics store has ten items of a given type in stock. One is defective. Four successive customers purchase one of the items. Each time, the selection is on anequally likely basis from those remaining. What is the probability that all four customes get good items?

SOLUTION

Let E i be the event the i th customer receives a good item. Then the first chooses one of the nine out of ten good ones, the second chooses one of the eight out of ninegoood ones, etc., so that

P ( E 1 E 2 E 3 E 4 ) = P ( E 1 ) P ( E 2 | E 1 ) P ( E 3 | E 1 E 2 ) P ( E 4 | E 1 E 2 E 3 ) = 9 10 8 9 7 8 6 7 = 6 10

Note that this result could be determined by a combinatorial argument: under the assumptions, each combination of four of ten is equally likely; the number of combinationsof four good ones is the number of combinations of four of the nine. Hence

P ( E 1 E 2 E 3 E 4 ) = C ( 9 , 4 ) C ( 10 , 4 ) = 126 210 = 3 / 5
Got questions? Get instant answers now!

A selection problem

Three items are to be selected (on an equally likely basis at each step) from ten, two of which are defective. Determine the probability that the first and third selected aregood.

SOLUTION

Let G i , 1 i 3 be the event the i th unit selected is good. Then G 1 G 3 = G 1 G 2 G 3 G 1 G 2 c G 3 . By the product rule

P ( G 1 G 3 ) = P ( G 1 ) P ( G 2 | G 1 ) P ( G 3 | G 1 G 2 ) + P ( G 1 ) P ( G 2 c | G 1 ) P ( G 3 | G 1 G 2 c ) = 8 10 7 9 6 8 + 8 10 2 9 7 8 = 28 45 0 . 62
Got questions? Get instant answers now!

(CP2) Law of total probability Suppose the class { A i : 1 i n } of events is mutually exclusive and every outcome in E is in one of these events. Thus, E = A 1 E A 2 E A n E , a disjoint union. Then

P ( E ) = P ( E | A 1 ) P ( A 1 ) + P ( E | A 2 ) P ( A 2 ) + + P ( E | A n ) P ( A n )

A compound experiment.

Five cards are numbered one through five. A two-step selection procedure is carried out as follows.

  1. Three cards are selected without replacement, on an equally likely basis.
    • If card 1 is drawn, the other two are put in a box
    • If card 1 is not drawn, all three are put in a box
  2. One of cards in the box is drawn on an equally likely basis (from either two or three)

Let A i be the event the i th card is drawn on the first selection and let B i be the event the card numbered i is drawn on the second selection (from the box). Determine P ( B 5 ) , P ( A 1 B 5 ) , and P ( A 1 | B 5 ) .

SOLUTION

From [link] , we have P ( A i ) = 6 / 10 and P ( A i A j ) = 3 / 10 . This implies

P ( A i A j c ) = P ( A i ) - P ( A i A j ) = 3 / 10

Now we can draw card five on the second selection only if it is selected on the first drawing, so that B 5 A 5 . Also A 5 = A 1 A 5 A 1 c A 5 . We therefore have B 5 = B 5 A 5 = B 5 A 1 A 5 B 5 A 1 c A 5 . By the law of total probability (CP2) ,

P ( B 5 ) = P ( B 5 | A 1 A 5 ) P ( A 1 A 5 ) + P ( B 5 | A 1 c A 5 ) P ( A 1 c A 5 ) = 1 2 3 10 + 1 3 3 10 = 1 4

Also, since A 1 B 5 = A 1 A 5 B 5 ,

P ( A 1 B 5 ) = P ( A 1 A 5 B 5 ) = P ( A 1 A 5 ) P ( B 5 | A 1 A 5 ) = 3 10 1 2 = 3 20

We thus have

P ( A 1 | B 5 ) = 3 / 20 5 / 20 = 6 10 = P ( A 1 )

Occurrence of event B 1 has no affect on the likelihood of the occurrence of A 1 . This condition is examined more thoroughly in the chapter on "Independence of Events" .

Got questions? Get instant answers now!

Often in applications data lead to conditioning with respect to an event but the problem calls for “conditioning in the opposite direction.”

Questions & Answers

a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
how did I we'll learn this
Noor Reply
f(x)= 2|x+5| find f(-6)
Prince Reply
f(n)= 2n + 1
Samantha Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive
Samson Reply

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Applied probability' conversation and receive update notifications?

Ask