Probability

Two events are mutually exclusive if they both cannot occur together. For example, if you toss a coin , let A represent a head showing up and B represent a tail showing up. These two events are mutually exclusive. You can only have a tail or head.

To explain an independent event, pick a card from a deck of 52. The probability that it is a king is 4/52. If you put the card back and draw again, the probability is still 4/52. The second draw is independent of the first draw. If you draw another card without putting it back, its probability changes to 3/51. It becomes a dependent event.

In short, a mutually exclusive event is not an independent event.

Rolling 2 dice together can produce total scores from 2 (a 1 and a 1) to 12 a (6 and a 6) but there not all equally probable. You would score 6 with 1 + 5 or 5 + 1 or 2 + 4 or 4 + 2 or 3 + 3.

Because there are 6 numbers on No. 1 dice and 6 numbers on No. 2 dice there are 6 x 6 = 36 ways of scoring.

From the above you can see that your 5 possibilities out of 36 means that the probability is 5/36 = 0.138888(recurring decimal) = 13.89% correct to 2 decimal places.

There is only one card which has 10 hearts upon it, and that is the 10 of hearts. Similarly there is only one two of clubs in a 52 pack.

Samplig frame is the source material from which the sample is drawn. If you have a 'list' of names of all inviduals from which you could draw a sample, the list is a sampling frame.

A samplig unit is the sample being chosen.

There are 15,000 milliliters in 15 liters.

land on a 3? as in a dice? well if so then it is 1/6 chance because there are 6 numbers on a dice and 3 is one of them so 1/6

Three: they are 3, 5 and 6.

If I have understood the question correctly, despite your challenging spelling, the standard deviation is the square root of the average of the squared deviations while the mean absolute deviation is the average of the deviation.

One consequence of this difference is that a large deviation affects the standard deviation more than it affects the mean absolute deviation.

The requirement that one coin is a head is superfluous and does not matter.

The simplified question is "what is the probability of obtaining exactly six heads in seven flips of a coin?"...

There are 128 permutations (27) of seven coins, or seven flips of one coin. Of these, there are seven permutations where there are exactly six heads, i.e. where there is only one tail.

The probability, then, of tossing six heads in seven coin tosses is 7 in 128, or 0.0546875.

The mean is 0 and the variance is 1. This need not be the case in any other Normal (Gaussian) distribution.

There are an indeterminate number: thousands upon thousands.

you have a 1/13 chance of getting an ace in a pack of cards

The different ways the vowels can be arranged together is 3x2x1 = 6.

The number of places this arrangement can go within the word is 5.

This means there are 30 possible arrangements for the vowels. The consonants can be arranged amongst themselves with 4 possibilities for first place, 3 for second, 2 for third and 1 for fourth. This gives us 4x3x2=24 combinations.

So with 30 possible arrangements for the vowels, and 24 for the consonants, this gives us 720 possible ways of arranging the letters in the word offices so that the vowels always come together.

If 10 coins are tossed, you could get 4 heads and 6 tails, 3 heads and 7 tails, 2 heads and 1 tail, 0 heads and 10 tails all giving fewer heads than tails.

Using the binomial distribution ,

P(4 heads) = 10C4 (.5)^4 (.5)^6 = 0.205078.

P(3 heads) = 10C3 (.5)^3 (.5)^7 = 0.117188

P(2 heads) = 10C2 (.5)^2 (.5)^8 = 0.043945

P(1 heads) = 10C1 (.5)^1 (.5)^9 = 0.009766

P(0 heads) =(.5)^10 = 0.000977

Adding all of these probabilities, we have P(fewer heads than tails)= 0.376953

It can be, if desired. Probability is a number between 0 and 1. If you want to express that as a percent, simply say between 0% and 100% instead.

The number of permutations of n distinct objects is n! = 1*2*3* ... *n.

If a set contains n objects, but k of them are identical (non-distinguishable), then the number of distinct permutations is n!/k!.

If the n objects contains j of them of one type, k of another, then there are n!/(j!*k!).

The above pattern can be extended. For example, to calculate the number of distinct permutations of the letters of "statistics":

Total number of letters: 10

Number of s: 3

Number of t: 3

Number of i: 2

So the answer is 10!/(3!*3!*2!) = 50400

1

depends what is the question asking

The number of 7 letter permutations of the word ALGEBRA is the same as the number of permutation of 7 things taken 7 at a time, which is 5040. However, since the letter A is duplicated once, you have to divide by 2 in order to find out the number of distinct permutations, which is 2520.

There are 13 clubs in a standard deck of 52 cards. The probability, then, of drawing club is 13 in 52, or 1 in 4, or 0.25.

A card is drawn from a standard deck of playing cards. what is the probability that a spade and a heart is selected?

There are 2 of the threes that meet your condition.

There are 12 face cards in a standard deck of 52 cards; the jacks, queens, and kings of spades, diamonds, clubs, and hearts. The probability, then, of drawing a face card is 12 in 52, or 3 in 13, or about 0.2308.

Yes, except that if you know that the distribution is uniform there is little point in using the empirical rule.

The number of ways that the letter of the word CANADA can be arranged is simply the number of permutations of 6 things taken 6 at a time, which is 6 factorial, or 720. However, since the letter A is repeated twice, the number is distinct permutations is a factor of 4 less than that, or 180.