Section 4.3: Conditional Probability Multiplication Rule

A fair coin is tossed five times. What is the probability that it comes up heads at least once?

#H: 0,1,2,3,4,5 P(0 heads)+P(at least 1 H)= 1 P(at least 1H)= 1-P(0 H)= 1-P(TTTTT)= 1-(1/2)^5= .9688

Multiplication rule for independent events

For any two independent events A and B, P(A and B) = P(A) x P(B) *simplified version of the multiplication rule

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?

P(<18 NYC)= .235 P(<18 Chi)= .258 P(<18 LA)= .26 P(NYC&Chi&LA)= .235 x .258 x .26= 0.016 P<0.5, so yes!

If A and B are independent events, then

P(A and B)= P(A) x P(B)

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?

P(detect flaw)=.8 # inspectors: 0,1,2,3 P(not detect)= .2 P(none detect)+P(at least 1 detect)= 1 P(at least 1 detect)= 1-P(none detect)= 1-P(not & not & not)= 1-(.2)^3= .992

An office has three smoke detectors. In case of fire, each detector has probability 0.9 of detecting it. If a fire occurs, what is the probability that at least one detector detects it?

P(detect)= .9 P(not detect)= .1 P(none detect)+P(at least 1 detect)= 1 P(at least 1 detect)= 1-P(none detect)= 1-(.1)^3= .999

In a certain city, 70% of high school students graduate. Of those who graduate, 40% attend college. Find the probability that a randomly selected high school student will attend college.

P(grad)=.7 P(coll | grad)= .4 P(grad & coll)= P(coll | grad) x P(grad)= .7 x .4= .28

My daughter just applied for PA school last spring. The probability of being granted an interview was 100/650. Among those interviewed, the probability of being offered a place in the class was 24/100. Find the probability that an applicant to her PA school was offered a spot in the class.

P(int)= 100/650= .1538 P(get in|int)= 24/100= .24 P(int & get in)= P(get in | int) x P(int)= .24 x .1538= .0369

A person is selected at random. • What is the probability that the person is a man? • What is the probability that the person is a man with a Bachelor's degree? • What is the probability that the person has a Bachelor's degree, given that he is a man?

P(man)= 94/195= .4821 P(man & bach)= 17.5/195= .0897 P(bach | man)= P(bach & man)/P(man)= .1861 or 17.5/94= .1862

Another way to find P(A & B):

We can also use the General Addition Rule to find this! ~If P(A or B) = P(A) + P(B) - P(A & B) we can rearrange this to find P(A & B)! ~P(A & B) = P(A) + P(B) - P(A or B)

Decide whether the sampled items can be treated as independent. a) A pollster plans to sample 1500 voters from a city in which there are 1 million voters. b)Five hundred students attend a college basketball game. Fifty of them are chosen at random to receive a free T-shirt.

a) .05(1,000,000)= 50,000 n=1500 < .05N so independent b) .05(500)= 25 n=50 is > .05N, so not independent

a) What is the probability that the person is a woman who is a high school graduate? b) What is the probability that the person is a high school graduate? c) What is the probability that the person is a woman, given that the person is a high school graduate?

a) P(w&HS)= 31.9/195= .1636 b) P(HS)= 29.6+31.9/195= .3154 c) P(W|HS)= P(W&HS)/P(HS)= .1636/.3154= .5187 or 31.9/ (29.6+31.9)= .5187

Approximately 15% of adult men in the U.S. are more than six feet tall. Therefore, if a man is selected at random, the probability that he is more than six feet tall is 0.15. Now assume that you learn that the selected man is a professional basketball player. With this extra information, the probability that the man is more than six feet tall becomes much greater than 0.15. A probability that is computed with the knowledge of additional information is called a

conditional probability. ~The conditional probability of an event B given an event A is denoted P(B|A). P(B|A) is the probability that B occurs, under the assumption that A occurs. *limits to a single row or a column, instead of the whole table ~The probability is computed as P(B|A)= P(AandB)/ P(A) *B given A

Sometimes we need to find the probability that an event occurs at least once in several independent trials. We can calculate such probabilities by

finding the probability of the complement and subtracting from 1. Remember that the complement of "at least one..." is "no events occur" P(A)+P(A^c)= 1 then subtract P(A) from both sides, you get P(A^c)= 1- P(A)

When sampling with replacement, the draws are __________________. When sampling without replacement, the draws are _______________.

independent; not independent *When a sample is very small compared to the population (less than 5%), a rule of thumb is that the items may be treated as independent n < .05N

Two dice are tossed. Let A be the event that the number on the first die is even, and let B be the event that the number on the 2nd die is 6. Explain why events A and B are independent. Find P(A), P(B) and P(A and B)

one die doesn't influence the other P(A)= P(even)= 3/6= .5 P(B)=P(6)= 1/6 P(1st even & 2nd 6)= 3/6 x 1/6= 3/36= .0833

Two events are independent if the occurrence of

one does not affect the probability that the other event occurs. ~If two events are not independent, we say they are dependent Example: A college student is chosen at random. The events are "being a freshman" and "being less than 20 years old." ~no, being a freshman makes it more likely to be <20yrs A college student is chosen at random. The events are "born on a Sunday" and "taking a statistics class." ~yes, no impact on each other

When we sample two items from a population, we can proceed in either of two ways. We can __________ the first item drawn before sampling the second. This is known as sampling with replacement. The other option is to __________________________ when sampling the second one. This is known as sampling without replacement.

replace; leave the item out

The General Multiplication Rule:

~The General Method for computing conditional probabilities provides a way to compute probabilities for events of the form "A and B." If we multiply both sides of the equation by P(A) we obtain the General Multiplication Rule. ~P(A and B) = P(A) * P(B|A) Or P(A and B)= P(B) x P(A|B) *if you have a table, just use the cell that gives "and"