Probability

There are twenty even-numbered card in a standard deck of 52 playing cards, the 2, 4, 6, 8, and 10 of Spades, Diamonds, Clubs, and Hearts.

The total number of days in a leap year is 366. Then, if we want to determine the probability of 53 Wednesdays occurring in a leap year, we write 53 / 366.

The probability of the coin coming up heads each time is 1/8; likewise for 3 tails. The probability of getting 2 heads and 1 tail (in any order) or 2 tails and 1 head, is 3/8. There are lots of other events whose probability can be calculated when a coin is tossed 3 times, but the question doesn't specify what event is to have its probability calculated.

1/6

Each leaf of the tree is one outcome. To take a simple example, consider the tree for three flips of a coin. The root splits into heads/tails, then each of the two branches splits into heads/tails, and finally each of the four branches splits into heads/tails, for 8 possible outcomes. In this simple case, each of the outcomes is equally probable. To find how many outcomes have exactly two heads, follow each branch of the tree, and put a star by the leaf each time you count two heads. You will find three leaves with exactly two heads (HHT, HTH, and THH). Add the probabilities of these three leaves, to discover that the probability of exactly three heads is 3/8.

You aren't close to providing enough information.

It is 4/52 or 1/13.

4 : Each coin toss multiplies the generated outcome by two. The outcome for two tosses can be Heads (H) Tails (T), H H, T T or T H. As such, with three tosses the outcome can be H H H, H H T, H T H, T H H, T T H, T T T, T H T, H T T.

We can define the number of outcomes as 2^n, where n is the amount of tosses.

1 in 4

If by 4 you mean either 3+1 or 2+2, then: (1/12)4(11/12)(5) = 55/(125) = 0.00022103

99/512, or 19.34%.

The nCr formula can be used in this case:

12!/((12-7!)*7!) ---> 95,040/120 ---> 792

792/(2^12) = 99/512

If p is the probability of the event happening (once), so that q = 1 - p is the probability of the event not happening (once), then the binomial theorem can be used:

(p + q)n = nC0 p0qn + nC1 p1qn-1 + ... + nCr prqn-r + ... + nCn-1 pn-1q1 + nCn pnq0

Where n is the number of trials and r is the number of successes required, and nCr is the number of combinations of r things from a set of n and is:

nCr = n!/(n-r)!r!

where n! is n factorial which is n x (n-1) x (n-2) x ... x 2 x 1. 0! is defined to be 1.

So assuming a normal, unbiased coin with p = pr(head) = 1/2 → q = pr(not_head) = 1 - 1/2 = 1/2, then the probability of getting (r =) 200 heads out of (n =) 1000 trials is:

nCr prqn-r = 1000C200 (1/2)200(1/2)800

= 1000!/(800! x 200!) (1/2)1000

Assuming they are standard dice numbers 1-6:

The minimum sum is 1 + 1 + 1 = 3

The maximum sum is 6 + 6 + = 18

So there are 18 - 3 + 1 = 16 different possible sums.

But note, that some of the sums may be obtained in more than one way, for example 4 can be obtained in 3 different ways: 1 + 1 + 2, 1 + 2 + 1, 2 + 1 + 1.

(4 + 4) in 52, or 8 in 52, or 2 in 13, or about 0.1538.

i dont know but why dont you do some research you lazy person

the first roll must be anything except a 1 .. odds = 5/6

the second roll must be one specific number that adds up to 8

so its odds are = 1/6

SO . . the total odds are 5/6 * 1/6 = 5/36 = about a 14% chance.

Another way to look at it is this:

there are only 5-combinatiosn that make 8 = 2/6, 3/5, 4/4, 5/3, 6/2

and there are 36 possible combinations . . .

so, again .. the odds are 5- out of 36 rolls.

If you assume that birth dates are uniformly distributed over the year (they are not), and you ignore leap years, then the probability of two people selected at random, share a birthday is 1/365.

{11,13,17,19,23,29} are all the prime numbers between 10 and 30

5/36

Out of the 36 possible and equally-probable outcomes (1st die's number and 2nd die's number), 5 combinations give you a sum equal 8:

* 1st die gives a 2 and 2nd die gives a 6

* 1st die gives a 6 and 2nd die gives a 2

* 1st die gives a 3 and 2nd die gives a 5

* 1st die gives a 5 and 2nd die gives a 3

* 1st die gives a 4 and 2nd die gives a 4

Therefore the probability that the sum is greater than 10 equals 5/36.

To calculate the number of zeros in a factorial number, we need to determine the number of factors of 5 in the factorial. In this case, we are looking at 10 to the power of 10 factorial. The number of factors of 5 in 10! is 2 (from 5 and 10). Therefore, the number of zeros in 10 to the power of 10 factorial would be 2.

Start with examples like flipping a coin, rolling a die or spinning a dreidel. Then explain in terms they understand. That depends very much on the age of the child.

It depends on the brand, it will be either 3 or 1.

They are used to understand how to set insurance premiums.

Take, for example, car insurance.

The probability of a new driver getting in an accident is higher than an experienced driver.

The probability of a male driver getting in an accident is higher than an female driver (in general).

The probability of a person who has had many claims filing another claim is higher than someone who has never filed a claim.

All these data guide the insurance industry to charge higher premiums to drivers who are new, male or have filed claims in the past.