Probability

There are (4*3)*(6*5*4*3*2*1) = 8640 ways.

4!(factorial) or 4x3x2x1 OR 24

There are 11 factorial, or 39,916,800 permutations of the letters in the word PROBABILITY. However, because the letters B and I are duplicated, there are only one quarter of that, or 9,979,200 distinguishable permutations.

true

There is no simple answer to the question because the children's genders are not independent events. They depend on the parents' ages and their genes.

However, if you assume that they are independent events then, given that the probability of a boy is approx 0.5169, the probability of the event described is 0.5169*1*1*0.4831 = 0.2497

When you throw a fair dice you have an equal possibility of getting 1, 2, 3, 4, 5 or 6. If you want to get a number greater than 4 this means you want 5 or 6: the probability is 2/6, which can be simplified to 1/3.

The mean value of 100, 70, 70, and 20 is 65

Population distribution refers to the patterns that a population creates as they spread within an area. A sampling distribution is a representative, random sample of that population.

There are 7 letters, but three 'o's are indistinguishable, so there are 7! ÷ 3! ways = 840 ways.

If the 'T' and the 't' are also considered to be indistinguishable (that is ignoring the case of the letters; making "Toronto" and "toronTo", etc the same), then the number of ways is 7! ÷ (3! x 2!) ways = 420 ways.

! after a number means the factorial of the number which is the product of all numbers less than or equal to the number, and greater than or equal to 1. Thus:

7! = 7 x 6 x 5 x 4 x 3 x 2 x 1.

3! = 3 x 2 x 1

2! = 2 x 1

0! is defined to be 1.

either 4 in 52 chances or 1 in 13 chances.

If only two cards are drawn from a standard deck of cards, with the first card replaced before drawing the second, the answer is 0.005917 (approx). If the first card is not replaced, the probability increases to 0.006033.

CHEESE

cheese has 6 letters in it. if you have a word with different letters, you do the factorial of the number of letters there are. but you can not do that in this word. since it has 3 E's, then you mut do 6!(factorial) divided by 3! then the answer would be 120

The answer depends on drawing a 9 from WHAT!

yes it is true

If the underlying distribution of the product is normally distributed then (and only then) the normal distribution can be used to identify specimens that are outside the acceptable range.

There are 12.

You calculate the standard error using the data.

There are 10 different sets of teachers which can be combined with 4 different sets of students, so 40 possible committees.

HHHH, HHHT, HHTT, HTTT, TTTT.

Pr(HHTT) = 6/16 = 0.375

The probability of tossing a coin twice and getting tails both times is 1 in 4, or 25%.

If you have already tossed a coin and had it land on tails, the probability that it will land on tails again the next time you toss it is 50%.

I'm assuming this is a combinations problem.

nCk = n!/(k!(n-k)!) is the general equation.

In your case n= 11 and k = 5

for 11!/(5!(11-5)!) = 11!/(5!*6!) = (11*10*9*8*7)/(5*4*3*2*1) = 462

The probability of an even must be a real number in the range [0, 1]. 120% is outside that range.

Two events are equally unlikely if the probability that they do not happen is the same for each event. And, since the probability of an event happening and not happening must add to 1, equally unlikely events are also equally likely,

Yes.

the mishap probability component is an estimation that a hazard will result in a mishap