5.3 (conditional probability and the multiplication rule
SOLUTION:P(comes up heads at least once=1-P(does not come up heads at all=1-P(comes up tails 5 times)=1-P(1st toss is T and 2nd toss is T.... and 5th toss is T) 1-(0.5)-(0.5)-(0.5)-(0.5)-(0.5)=0.96875
1-(1-% )^however many times it's done
EXAMPLE:A fair coin is tossed five times. What is the probability that it comes up heads at least once?
SOLUTION:P(comes up heads at least once=1-P(does not come up heads at all=1-P(comes up tails 5 times)=1-P(1st toss is T and 2nd toss is T.... and 5th toss is T) 1-(0.5)-(0.5)-(0.5)-(0.5)-(0.5)=0.96875
Remember that the complement of "At least one event occurs" is "No events occur".To compute the probability that an event occurs at least once, find the probability that _______ _____________________________________________________________________________.
Find probability it does not occur at all and subtract 1
EXAMPLE:According to recent figures from the U.S. Census Bureau, the percentage of people under the age of 18 was 23.5% in New York City, 25.8% in Chicago, and 26.0% in Los Angeles. If one person is selected from each city, what is the probability that all of them are under 18? Is this an unusual event?
Independent P(New York)P(Chicago)P(Los Angeles) =0.0158
THE MULTIPLICATION RU LE FOR INDEPENDENT EVENTS:
P(A and B)= P(A)*P(B)
If you have all independent events, solution is product of all 3 (multiply them)
P(A and B)=P(A)P(B)
The Conditional Probability an event B given an event A is denoted_______________. P(B/A) is the probability that B occurs, under the assumption that A occurs.
P(B/A)
Conditional Probability: The probability is computed as: P(B/A)=
P(B/A)= P(A and B) / P(A)
THE GENERAL MULTIPLICATION RULEThe General Method for computing conditional probabilities provides a way to compute probabilities for events of the form "Aand B." If we multiply both sides of the equation by P(A) we obtain the General Multiplication
Rule.THE GENERAL MULTIPLICATION RULE: P(A and B)=P(A)*P(B/A) P(A and B)=P(B)*P(A/B)
EXAMPLE:Items are inspected for flaws by three inspectors. If a flaw is present, each inspector will detect it with probability 0.8. The inspectors work independently. If an item has a flaw, what is the probability that at least one inspector detects it?
SOLUTION:
EXAMPLE:Consider the following table which presents the number of U.S. men and women (in millions) 25 years old and older who have attained various levels of education.Not a high school graduateHigh school graduateSome college, no degreeAssociate's degreeBachelor's degreeAdvanced degree Men14.029.615.67.217.510.1 Women13.731.917.59.619.29.1A person is selected at random. a)What is the probability that the person is a man?b)What is the probability that the person is a man with a Bachelor's degree?c)What is the probability that the person has a Bachelor's degree, given that he is a man?
SOLUTION: a)0.4821 b)0.08974 c) 17.5/195.0/94.0/195.0= 0.1862
EXAMPLE:Among those who apply for a particular job, the probability of being granted an interview is 0.1. Among those interviewed, the probability of being offered a job is 0.25. Find the probability that an applicant is offered a job.
SOLUTION: (0.1)(0.25) =0.025
"AT LEAST ONE"TYPE PROBABILITIESSometimes we need to find the probability that an event occurs at least once in several independent trials. The easiest way to calculate these probabilities is by finding the probability of the __________________________.
complement
A probability that is computed with the knowledge of additional information is called a ______________________________.
conditional probability
If two events are not independent, we say they are ____________________.
dependent
DEPENDENCE Two events are ____________________ if the occurrence of one does not affect the probability that the other event occurs.
independent