Probability

CHOICES

This is a non-random sampling technique where the initial respondents to a survey recruit others for a survey.

Linear snowball sampling is when the first responent recruits one more person for the survey, who in turn recruits one more person for the survey. Number of people surveyed will increase in a linear manner.

The related links provide good discussion of this form of surveying. It is a non-representative sample, but it is an effective means of surveying people with common traits who might otherwise be difficult to find.

For example, I want a statistical profile of people that are "hackers" so every hacker I find, I would ask if I could interview his "fellow hackers."

See related links.

1/4

There are four 7s in a deck of 52 cards. Therefore probability of picking a 7 is 4/52 or 1/13 or 0.0769.

It depends on the number of sides and how many times you roll it.

5/20, 25%, or .25

The sample space of a standard six sided die is [1,2,3,4,5,6].

1.14

P(a^~a)==1 P(a&~a)==0

the line above is shorthand notation for an event that has a probability of 1, followed by an event that has a probability of 0. P(event) is an easy way to say the probability of "event". The "^" means "OR", the "~" means "NOT", and the "&" as you are probably familiar means "AND". So puting it together, "P(a^~a)" means the probability that an event "a" occurs OR that event "a" does not occur. So take an event "a", any event, like drawing an 8 of clubs out of a deck of cards. If you draw a random card out of a deck of cards, the probability that you will draw an 8 of clubs OR that you will not draw an 8 of clubs is 1. That means 100%. So when you draw a card you will either draw the 8 of clubs or not draw the 8 of clubs. It seems like an obvious statement to make but its a proof that becomes very important in proving less obvious theories. likewise, the second statement was "P(a&~a)", so the probability that event "a" occurs AND event "a" doesnt occur. Since the event has to either occur or not, it cant occur AND not occur, so the probability is 0.

Less. The more times the coin is tossed, the more likely it will reflect the actual odds of .5 heads and .5 tails.

true

No, it is continuous.

On a fair 50-50 coin, the chance of you getting heads 3 times in a row is .5 * .5 * .5 which is 12.5%

Getting exactly 3 heads out of any number of coin flips involves:

(Number of Flips!/ [6 * (N-3)!]) * (.5^3)* (.5^(n-3))

It means that every member of the population has the same probability of being included in the sample.

It is 6.1, approx.

(1,1,1) (1,1,2) (1,1,3) (1,1,4) (1,1,5) (1,1,6) (1,2,1) (1,2,2) (1,2,3) (1,2,4) (1,2,5) (1,2,6) (1,3,1) (1,3,2) (1,3,3) (1,3,4) (1,3,5) (1,3,6) (1,4,1) (1,4,2) (1,4,3) (1,4,4) (1,4,5) (1,4,6) (1,5,1) (1,5,2) (1,5,3) (1,5,4) (1,5,5) (1,5,6) (1,6,1) (1,6,2) (1,6,3) (1,6,4) (1,6,5) (1,6,6) (2,1,1) (2,1,2) (2,1,3) (2,1,4) (2,1,5) (2,1,6) (2,2,1) (2,2,2) (2,2,3) (2,2,4) (2,2,5) (2,2,6) (2,3,1) (2,3,2) (2,3,3) (2,3,4) (2,3,5) (2,3,6) (2,4,1) (2,4,2) (2,4,3) (2,4,4) (2,4,5) (2,4,6) (2,5,1) (2,5,2) (2,5,3) (2,5,4) (2,5,5) (2,5,6) (2,6,1) (2,6,2) (2,6,3) (2,6,4) (2,6,5) (2,6,6) (3,1,1) (3,1,2) (3,1,3) (3,1,4) (3,1,5) (3,1,6) (3,2,1) (3,2,2) (3,2,3) (3,2,4) (3,2,5) (3,2,6) (3,3,1) (3,3,2) (3,3,3) (3,3,4) (3,3,5) (3,3,6) (3,4,1) (3,4,2) (3,4,3) (3,4,4) (3,4,5) (3,4,6) (3,5,1) (3,5,2) (3,5,3) (3,5,4) (3,5,5) (3,5,6) (3,6,1) (3,6,2) (3,6,3) (3,6,4) (3,6,5) (3,6,6) (4,1,1) (4,1,2) (4,1,3) (4,1,4) (4,1,5) (4,1,6) (4,2,1) (4,2,2) (4,2,3) (4,2,4) (4,2,5) (4,2,6) (4,3,1) (4,3,2) (4,3,3) (4,3,4) (4,3,5) (4,3,6) (4,4,1) (4,4,2) (4,4,3) (4,4,4) (4,4,5) (4,4,6) (4,5,1) (4,5,2) (4,5,3) (4,5,4) (4,5,5) (4,5,6) (4,6,1) (4,6,2) (4,6,3) (4,6,4) (4,6,5) (4,6,6) (5,1,1) (5,1,2) (5,1,3) (5,1,4) (5,1,5) (5,1,6) (5,2,1) (5,2,2) (5,2,3) (5,2,4) (5,2,5) (5,2,6) (5,3,1) (5,3,2) (5,3,3) (5,3,4) (5,3,5) (5,3,6) (5,4,1) (5,4,2) (5,4,3) (5,4,4) (5,4,5) (5,4,6) (5,5,1) (5,5,2) (5,5,3) (5,5,4) (5,5,5) (5,5,6) (5,6,1) (5,6,2) (5,6,3) (5,6,4) (5,6,5) (5,6,6) (6,1,1) (6,1,2) (6,1,3) (6,1,4) (6,1,5) (6,1,6) (6,2,1) (6,2,2) (6,2,3) (6,2,4) (6,2,5) (6,2,6) (6,3,1) (6,3,2) (6,3,3) (6,3,4) (6,3,5) (6,3,6) (6,4,1) (6,4,2) (6,4,3) (6,4,4) (6,4,5) (6,4,6) (6,5,1) (6,5,2) (6,5,3) (6,5,4) (6,5,5) (6,5,6) (6,6,1) (6,6,2) (6,6,3) (6,6,4) (6,6,5) (6,6,6)

Yes.

0

Prob(Exactly 9 Heads) = 0.0269

Prob(At least 2 Heads) = 0.0327

Prob(At most 8 Heads) = 0.9673

Associates a particulare probability of occurrence with each outcome in the sample space.

Probability

18

The probability of selecting 1 female at random is 4/8

There are now 7 students, 3 of which are female so the probability of selecting another female is 3/7

The probability two randomly selected students are female is (4/8)x(3/7)= 3/14

The probably of four girls in a family with four children is 1/16. I got this answer because:

Probability of a girl is 1/2 and to get all girls you would multiply it by 1/2 for the rest of the girls.