Probability

A fair coin means that the probability of a head = probability of a tail = 1/2

So you would expect half the tosses to be heads, ie 1/2 x 75 = 371/2 heads.

...oooOOOooo...

Having 1/2 a head doesn't seem possible, but when the question asks about expectation, it is saying: if you repeated the experiment lots of times, how often, on average, would the required result appear.

So the expectation of heads when a fair coin is tossed 75 times is asking: if a fair coin was repeatedly tossed 75 times, what would be the (mean) average number of heads achieved?

As more and more trials are done and the (mean) average of the number of heads got is taken, it will get closer and closer to 371/2

37 or 38 times. (Obviously, you can't have half of a time.) You will either get one or the other, and a fair coin means that either is just as likely. So, it should split evenly down the middle.

Random selection.

Gwerth Probablitiy

There are 4 Kings in a standard pack of 52 cards.

If 1 King has previously been drawn this now leaves 3 kings out of a total of 51 remaining cards. The probability of now drawing a King is therefore 3/51 which simplifies to 1/17.

Note: this is the probability concerning the 2nd draw only.

1/100

There are eight different letters in 'mathematics' - a, c, e, h, i, m, s and t.

Since there are 26 letters in the alphabet, the chance of a randomly chosen one being a letter that's in mathematics is 8/26, or 4/13.

The probability of rolling a odd number on a standard die is 3 in 6, or 1 in 2, or 0.5.
You don't say with what, so I assume with a standard six sided die. The probability of rolling an odd number with a standard six sided die is 3 in 6, or 1 in 2, or 0.5.

2600

The chance of heads the first time is 50%.
The chance of heads the second time is 50%.
The chance of heads the third time is 50%.
The chance of heads the fourth time is 50%.

The chance of all four is (0.5 x 0.5 x 0.5 x 0.5) = 0.0625 = 6.25 percent

Since there are two permutations of two dice that sum to 11, (5-6) and (6-5), and since there are 36 permutations of rolling two dice, the probability of not rolling a sum of 11 is 34 in 36, or 17 in 18, or about 0.9444.

me!!!

each 10 times they use the same numbers!!!!

here are mine:

7 - 42 - 36 - 21 - 11 - 32 - 5

use these for the next lotto!!!

you will win £1,000,000,000,000,000

24 ways.

It is called ONE = 1.

4 digit number having 0 at last position = 9*8*7

=504

4 digit even number having 2 or 4 or 6 or 8 at last position = 8*8*7*4

=1792

(in first position, 0 cannot be fix and in last position one of digit 2,4,6,8 will be fix,

so, number of choice left for first position is 8)

So, total 4 digit even number possible = 504 + 1792

= 2296

7/8

P(all heads) = (1/2)*(1/2)*(1/2)=(1/8).

P(at least one tails) = 1-P(all heads) = 7/8.

Another (more time-consuming) way to get the same result:

P(all tails) = (1/2)^3=1/8

P(exactly two tails) = 3*(1/2)^3=3/8

P(exactly one tails) = 3*(1/2)^3=3/8

P(at least one tail) = P(all tails) + P(exactly two tails) + P(exactly one tails) = 7/8

There is a 50% chance that it will land on heads each toss. You need to clarify the question: do you mean what is the probability that it will land on heads at least once, exactly once, all five times?

The probability of a coin landing head-side down is 0.5

The probability of landing head-side up is 0.5

Did you mean to ask the probability of it landing on its edge ?

You find the mean of negative numbers the same way you find the mean of any numbers. You add them up and then you divide by the number of numbers.

An example, using -3, -4, -6, and -9. The sum is -22. N is 4. The mean is -5.5.

The probability of this is 50%. 2/4

True.

Only 6 is greater than 5 on a standard cubic die. The chance of rolling a 6 is one in six.

The first die can come up in any one of 6 ways.
The second die can come up in any one of 6 ways.
There are (6 x 6) = 36 different ways that two dice can fall.

There are 5 ways that they can come up as a six:

1 + 5
5 + 1

2 + 4
4 + 2

3 + 3

The probability of rolling a 6 is (5/36) = 13.89 %

The odds in favor are 5 to 31.

1.96

Tackling multiple choice questions relies primarily on the memory process called recognizing.

George Wickham did not attend the ball at Netherfield, because he did not want to meet Mr. Darcy