Probability

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

If X has any discrete probability distribution then the sum of a number of observations for X will be normal.

yes, but only with memorization and scripts

Mean: 26.33

Median: 29.5

Mode: 10, 35

Standard Deviation: 14.1515

Standard Error: 5.7773

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.

Independent events.

The larger the sample size, the smaller the margin of error.

This is because the normal distribution has a domain that extends to infinity in both directions.

There are four different ways that two coins can land:

T T

T H

H T

H H.

Only one of them is two heads, so if the coins are fair, then

the probability is 1/4 = 25% .

There are 52 cards in a deck

there are 4 aces and 4 kings which makes a total of 8 kings and aces.

Assuming that the deck is full and shuffled the probability of drawing an aces or a king is 8/52 which simplifies to 2/13

Assuming that the arrays need to be rectangular, the answer is 14,597,412,049,059,800,000 or approx 15 quintillion.

z = (75 - 85)/5 = -10/5 = -2

no of possibilities

for example tossind a fair coin then

the cardinality of sample space is 2

It is 0.0606

z = 0.8416

4/6 = 2/3

Because the probability of an impossible event is 0 and probability lies within the area of {0-1} meaning 0 is impossible and 1 is definite. If you took an improper fraction such as 5/4 and made it a proper faction, it would be 1 1/4 making it more than 1 or (100%) and that wouldn't make any sense because 4/4 (1) is definite, anything more isn't a probability.

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There need not be any. Most decks of cards have backs that are geometric patterns, not eyes.

They are both measures of the likelihood of specified events.

Because as the sample size increases the Student's t-distribution approaches the standard normal.

The probability of pulling a four of diamonds from a standard deck of playing cards with the jokers removed is 1:52, the same as pulling any other individual card from the deck.