Probability

{hhhhhh, hhhhht, hhhhth, hhhhtt, hhhthh, hhhtht, hhhtth, hhhttt, hhthhh, hhthht, hhthth, hhthtt, hhtthh, hhttht, hhttth, hhtttt, hthhhh, hthhht, hthhth, hthhtt, hththh, hththt, hthtth, hthttt, htthhh, htthht, htthth, htthtt, httthh, htttht, htttth, httttt, thhhhh, thhhht, thhhth, thhhtt, thhthh, thhtht, thhtth, thhttt, ththhh, ththht, ththth, ththtt, thtthh, thttht, thttth, thtttt, tthhhh, tthhht, tthhth, tthhtt, tththh, tththt, tthtth, tthttt, ttthhh, ttthht, ttthth, ttthtt, tttthh, ttttht, ttttth, tttttt}

Edit: It removed all my capitalization, but everything should be capitalized.

This problem doesn't lead to a whole number solution. Are you sure you copied it correctly? Maybe you meant the ratio is 4 to 3. In that case:

red/blue = 4/3

blue/red = 3/4

blue = (3 * red) / 4 = (3 * 16) / 4 = 12 blue marbles

Consider one placement at a time. The first book on the shelf could be any of seven. The second could be any of the remaining six. This continues with the rest. The total combinations are:

7! = 7*6*5*4*3*2*1 = 5040

Nearly true. It is a point estimate, not point ofestimate.

When, over a given range, the probability that a variable in question lies within a particulat interval is equal to the size of that interval as a proportion of the range.

Generally, the larger the sample the more reliable the results.

Example:

If you flipped a coin twice and got heads both times you could say the coined is biased towards heads.

However, if you repeat the experiment 100 times your results will be a lot more reliable.

Population Parameter

When the odds are 5 to 2, if you bet $2 on something and win then you will receive $7 The $2 you bet and the $5 you won.

drawing a club

total club =13

P(Club)= 13/52

drawing a 7

total 7 = 4

P(7) = 4/52

P(a club and a 7) = P(Club) + P(7) as events are mutualy exculisve

P(a club and a 7) = 13/52+4/52

P(a club and a 7) = 17/52

Since there are 6 numbers on a die (1-6), then the probability of rolling a 5 would be 1 out of 6.

The probability of having a girl versus a boy is 1/2 because there is two things you have a chance of getting and you can only get one or the other.

The discrete random variable almost always arises in connection with counting.

The probability of the event occurring.

12/20, 3/5, or 0.6

The probability is the number of girls divided by the number of students, so 12/22, or 6/11

Number of useful outcomes over number of possible outcomes and simplify it if you can. Imagine you want an even number and you roll a die. There are 6 possible outcomes and three of them are useful outcomes (outcomes we want).

3

6

Simplify it and you get

1

2

13 out of 52

2/9

3/5=g/30

You got a one or a four. One is the zeroeth multiple of 4. Four is the first multiple of four. Any other result, such as 8, and you are not rolling a standard die.

ok I'm sure not many people will be asking this but a biased spinner is basically a spinner that is unfair or not accurate

5/6 0.833333333 ect. 83%

It's the same as a z-Transformation.

for all xi: (xi-mean(x)) / std(x)

If there is zero deviation all the observations are 50.

The probability of at least 1 match is equivalent to 1 minus the probability of there being no matches.

The first person's birthday can fall on any day without a match, so the probability of no matches in a group of 1 is 365/365 = 1.

The second person's birthday must also fall on a free day, the probability of which is 364/365

The probability of the third person also falling on a free day is 363/365, which we must multiply by the probability of the second person's birthday being free as this must also happen. So for a group of 3 the probability of no clashes is (363*364)/(365*365).

Continuing this way, the probability of no matches in a group of 41 is (365*364*363*...326*325)/36541

This can also be written 365!/(324!*36541)

Which comes to 0.09685...

Therefore the probability of at least one match is 1 - 0.09685 = 0.9032

So the probability of at least one match is roughly 90%