Probability

You think of each toss as having heads or tails, so there are two choices. If you toss the coin twice you have 4 choices HH, TT, HT and TH. The number of different coin tosses is 2n ordered tosses and n+1 unordered. For example in two tosses, 22=4 ordered tosses if HT is different than TH and if HT is the same as TH then we have 2+1=3 different possible tosses. So using this if we toss a coin 5 times and order does not matter, there are 6 different possibilities. These are HHHHH, TTTTT, HTTTT, HHTTT, HHHTT, HHHHT,HHHHH.

If order matters, there are 25=32 different possible ways to toss a coin 5 times. For example, HHHHT is different than THHHH. I will not write them all out, but you can do it easily by starting with HHHHH and changing each H to a T then two H to T and then 3, etc.

how do we analyze favorable and possible outcomes

rising action

none

There are 4 queens in 52 cards so the odds of picking a queen in one draw if 4/52. Now once you have picked a queen, there are 3 queens left and 51 cards. So 4/52 x 3/51 = 1/221 = 0.45 %

20, 40, 60, 80, or 100.

The probability is 0.2503

24 or 16

It is 1 - prob(none of the four is a heart)

= 1 - (39/52)*(38/51)*(37/50)*(36/49)

= 1 - 1974024/6497400 = 0.6962

atomic orbital

We'd like to know the likelihood that any 5 people chosen from 20 will be left-handed if the probability of being left-handed is assumed to be 10%. Well, 5 people can be chosen from 20 in (20 C 5) = 15504 ways. So, we have (15504)*(.1^5)*(.9^15) = 3.19%. This is a Binomial probability problem, where we are interested in 'k' successes (k = 5, success --> people being left-handed) in 'n' trials (n = 20).

The Poisson distribution is discrete.

False. It is approximately 1. Theoretically, it is not 1. I used excel, and I know the probability is between 0.999999 and 1. as the probability of Z<6 is 0.999999. I can't calculate the probability exactly because excel only goes to 7 place accuracy.

It should be a 16% chance. If she makes 2 out of 5 free throws, that's a 40% chance she'll make a free throw. The probability that she makes two in a row is therefore .4 * .4 = .16, or 16%.

P( a student getting an A) = 5/20=1/4

There are 3 students. The probability that all three got an A is (1/4)(1/4)(1/4)=1/64.

hypergeometric distribution f(k;N,n,m) = f(3;52,4,3)

There are 32 possible outcomes.

HHHHH,

HHHHT, HHHTH, HHTHH, HTHHH, THHHH,

HHHTT, HHTHT, HHTTH, HTHHT, HTHTH, HTTHH, THHHT, THHTH, THTHH, TTHHH,

HHTTT, HTHTT, HTTHT, HTTTH, THHTT, THTHT, THTTH, TTHHT, TTHTH, TTTHH,

HTTTT, THTTT, TTHTT, TTTHT, TTTTH,

TTTTT.

The answer is 7!/5! = 42 ways.

No, it resembles a normal distribution, but discrete.

sum --- Probability

2-----------1/36

3-----------2/36

4-----------3/36

5-----------4/26

6-----------5/36

7-----------6/36

8-----------5/36

9-----------4/36

10----------3/36

11----------2/36

12----------1/36

hypergeometric distribution f(k;N,n,m) = f(1;51,3,1)

or binominal distribution f(k;n,p) = f(1;1,3/51) would result in same probability

This question does not have a unique answer unless we make an additional assumption: we must know something about the relationship between the trials. For example, if we are monitoring a lightbulb for failure and the "event" consists of a working lightbulb, then the event not occurring means the lightbulb doesn't work. Let the "trials" consist of turning on the lightbulb. If the bulb fails on the first trial (chance of 1 - 3/8 = 5/8) then obviously it will not work on the second trial, either. The answer in this case would be 5/8 = 62.5%. If we assume the two trials are statistically independent, we are really saying we don't want to worry about these issues. Equivalently, the person asking the question is just telling us to multiply the probabilities (which means they are probably a teacher or a textbook and they or it are mainly concerned about your ability to multiply fractions, not about your understanding of probability). The chances of the event not occurring in each trial are 5/8. Multiplying gives 25/64 = about 39%. This is quite a bit less than 62.5%.

there is a 1/2 chance that it will land on heads and 1/2 chance it will land on tails. it dosent matter what the stats are

The probability is 0.0035

There are 6*6*2 = 72 possible outcomes.