Probability

The answer will depend on what "gr" is.

The answer will depend on what "gr" is.

The answer will depend on what "gr" is.

The answer will depend on what "gr" is.

sample space = 1, 2, 3, 4

The probability of getting an odd number in a single throw of a fair die (not dice!) is 1/2.

The probability of getting an odd number in a single throw of a fair die (not dice!) is 1/2.

The probability of getting an odd number in a single throw of a fair die (not dice!) is 1/2.

The probability of getting an odd number in a single throw of a fair die (not dice!) is 1/2.

8 four 4's and four 9's

unless you want to count aces as 1 then there would be 12

The probability of 4 aces being in a hand of 9 cards is:

9C4 ∙ (4/52)∙(3/51)∙(2/50)∙(1/49)∙(48/48)∙(47/47)∙∙∙(44/44) = 0.0004654...

≈ 0.0465%

where 9C4 = 9!/[(9-3)!∙3!] = 126

The answer depends on how many dice are rolled. With 12 or more dice, the answer is 0.

The answer depends on how many dice are rolled. With 12 or more dice, the answer is 0.

The answer depends on how many dice are rolled. With 12 or more dice, the answer is 0.

The answer depends on how many dice are rolled. With 12 or more dice, the answer is 0.

6

The range of vales of the event is the outcome space.

If you roll the die often enough, the event is a certainty and so the probability is 1.

If you consider only the first 5 rolls, the answer is (1/6)5 = 1/7776 = 0.000129 approx.

There are 720.

The answer depends on how often the die is rolled. As the number of rolls increases, the probability gets near enough to 1 as makes no difference.

For a single roll, the answer is 1/3.

The answer depends on how many cards are drawn, whether that is with or without replacement, whether the cards are drawn at random. If only one card is drawn, the probability is 0. If 51 cards are drawn, the probability is 1.

If two cards are drawn, at random, and the first is not replaced, the probability is

(2/52)*(1/51) = 2/2652 = 0.00075, approx.

This is easiest calculated by calculating the probability that NO SINGLE heads is obtained; this is of course the complement of the question. The probability of this is 1/2 x 1/2 x 1/2 ... 7 times, in other words, (1/2)7. The complement, the probability that at least one head is obtained, is then of course 1 - (1/2)7, or a bit over 99%.

We would need to know which events you're referring to to respond to this question.

This is the same as the probability of choosing either a red of a blue marble.

There are 5+4 out of 15 ways of doing this. The probability is therefore 9/15 = 3/5.

There are only 2 outcomes in which this happens. Di 1 = 5, Di 2 = 6 or Di 1 = 6, Di 2 = 5.

This is 2 possibilities out of 36. 2/36 is 0.055556.

There are many methods: stratified random sampling and cluster sampling are two examples.

Suppose you have a school with 1000 pupil: 400 in the Junior school and 600 in the Senior school. You may wish to keep the sampling proportions in the two parts of the school the same. So if you wanted a sample of 5% = 50 pupils, you would take a probability or random sample of 5% = 20 from the Juniors and 5% = 30 from the Seniors.

For the second example, imagine you want to sample 5% of all schools in the country. This could result in you spending lots of time and money travelling from place to place. Instead, you divide up the country into 100 regions - each containing the same number of schools. You then take a probability sample of 5% of these regions. In each of the regions you visit every school.

1/2 x 1/2 = 1/4

A dependent probability.

Theoretical probability = 0.5

Experimental probability = 20% more = 0.6

In 50 tosses, that would imply 30 heads.

The probability of not getting a club is the same as the probability of getting one of the other suits. There are (52-13)=39 such possibilities out of 52. Hence the probability is 39/52=3/4.

The answer depends on how many cards are drawn.

Although I cannot prove it, I doubt that there is a nice closed-form expression for this quantity; it's probably not a rational number.

It's value is approximately 0.0227501319482. (Using Python code.)

These are sometimes called 'electron clouds'.

On a fair normal die it is 1/2.