Probability

The number of distinct arrangements of the letters of the word BOXING is the same as the number of permutations of 6 things taken 6 at a time. This is 6 factorial, which is 720. Since there are no duplicated letters in the word, there is no need to divide by any factor.

In the normal distribution, the mean and median coincide,

and 50% of the data are below the mean.

Once...its true!

Not possible to tell you without knowing how many students' there are, and what distribution you wish to use (i.e normal distribution, t-distribution etc...)

the probability is 1(being the maximum)- the probability you have allredy got.

the answer is 0.6

there's 1/3 chance of getting an even number in a die, hon

The probability of identical repetition is 0 since there are infinitely many positions and orientations for each piece in the kaleidoscope.

No. The probability of an event ranges from 0 (the event will not happen) to 1 (the event will happen).

If you don't replace the cards then it's:

(4/52 x 3/51 x 2/50 x 1/49)

Couldn't be bothered working it out

There are 5 letters to pick for the first spot in our new arranged word, 4 letters in the 2nd spot 3 for the 3rd spot 2 for the fourth spot and only 1 for the last spot. multiple those together (5*4*3*2*1) but because we have duplicate letters we have to divide by our number of duplicates (2) so our final number is 120/2 or 60 ways hope this helps =D

z = (x - μ) / σ is the formula where x is the raw score and z is the z-score. μ and σ are the mean and standard deviations and must be known numbers.

Multiply both sides by σ

zσ = x-μ

Add μ to both sides

μ + zσ = x

x = μ + zσ

You calculate the raw score x , given the z-score, μ and σ by using the above formula.

1 in 2 to get tails on first toss

1 in 2 to get heads on second toss

To get tails on first toss and heads on second toss, probability is

1/2 x 1/2 = 1 in 4

As there are 4 queens in a 52 cards deck the probability is P=4/52=1/13

Simple random sampling.

When you roll two dice, a 7 may occur as (1,6),(2,5),(3,4),(4,3),(5,2),(6,1)

There are 6 elements in the event space.

If they are normal dice, the probability is 0.

yes?

To do this, you first need to convert the percentage into a z-score. The bottom 10% yields a z-score of -1.2816. Multiplying this by 55 and adding to the mean gives 69.512. This means all score less that are 69 or less will be in the bottom 10%

1/6

The probability of rolling any specific number will be 1/6, as there are 6 equally likely numbers on a dice. The probability of rolling a 6 then a 4 will thus be 1/6 x 1/6 which gives 1/36. Note, however, that there are two ways the dice can be ordered. The first can be 6 and the second 4, or the other way around. So multiply the above probability by 2 and you get 1/18.

Thus the probability of rolling a 6 and a 4 is 1/18

The odds of drawing a queen is 4 in 52, or 1 in 13. The odds of drawing a heart is 13 in 52, or 1 in 4. The odds, then, of drawing a queen or a heart is (4 + 13 - 1) in 52, or 16 in 52, or 4 in 13. (The reason for subtracting one is to account for the queen of hearts, which is both a queen and a heart.)

To find out the probability of something like this, you just find the number of diamonds and divide it by the total number of cards. In this case, there are 13 diamonds in a deck, and 52 cards. Dividing 13 by 52 gives 0.25. This can then be converted into a percentage, making the probability 25% or into a fraction, making the probability 1/4

queens and princes

First work out the probability of the first two coins being heads and then the last being tails. This is 1/2 x 1/2 x 1/2 which is 1/8

The next step is to find out how many different orders the coins can come in. In this case there are 3 possible orders (HHT, HTH, and THH). Multiply this by the above probability and you get 3/8. Therefore the probability of getting two heads and one tail is 3/8

The answer depends on the underlying variance (standard deviation) in the population, the size of the sample and the procedure used to select the sample.