The probability of rolling 4 and 4 on a single roll of two dice is 1 in 36. You don't have to count. The probability of rolling 4 on one die is 1 in 6. The probability of rolling two 4's is (1 in 6) squared or 1 in 36. If you insist on counting, go ahead, 11, 12, 13, 14, 15, 16, 21, 22, 23, 24, etc. You will find 36 combinations, with only one 44 in them, hence 1 in 36.
Why diamond king in playing cards has one eye?
He is supposed to be the king ceaser. who was one eyed. He had lost one eye during the war.
What is the probability of pulling 7 spades in a row from a deck of 52 cards?
The answer is 8,648,640 in 674,274,182,400 or approximately 0.128%.
To pull seven spades in a row from a deck of 52 cards may be translated as follows:
In a deck of 52 cards there are 13 spades, so 13 chances to pull a spade or 13/52.
After pulling one spade, there are 12 spades left and 51 cards left, so 12 chances to pull a spade or 12 / 51.
After pulling the second spade, there are 11 spades left and 50 cards left, so 11 chances to pull a spade or 11 / 50.
Continuing the above logic to the natural conclusion, we get the following multipliers:
13 / 52 * 12 / 51 * 11 / 50 * 10 / 49 * 9 / 48 * 8 / 47 * 7 / 46
which is equal to 8,648,640 / 674,274,182,400 or approximately 0.128%.
What is the probability of rolling a six sided die 20 times and getting a 5?
Suppose A is the event {rolling a 6-sided die 20 times and getting a 5}
Then A', the complement of A, is the event {rolling a 6-sided die 20 times and not getting a 5 any times}
So Pr(A) = 1 - Pr(A') = 1 - (5/6)20 = 1 - 0.026084 = 0.973916 (to 6 dp)
What is the probability to have 2 girls and a boy in a family?
The 8 possible outcomes for three children are:
* ggg
* ggb
* gbg
* gbb
* bgg
* bgb
* bbg
* bbb
Of these, two girls and a boy occurs 3 out of 8 times, which is a probability of 0.375. This assumes that the probability of a boy and girl being in the family is equal, which is not entirely true for a large number of reasons.
The odds of getting 4 cards in any particular suit (for example, hearts) would be:
1/4*1/4*1/4*1/4=1/256.
The odds of getting 4 cards of the same suit (any suit) would be 4x greater:
1/256*4=1/64
Because it doesn't matter what the last card is, the probability is 1/64.
Since 48 cards of a 52-card deck are not aces, you are guaranteed to have at least two aces only if you have at least 50 cards.
Since there are four suits for each rank, you are guaranteed to have at least two ranks whenever you have at least five cards.
Everything about the m-1 garand?
Scott Duff has a book that is very good.
Everything is a lot to ask for on the computer, so I suggest you follow the above advice and read the book, but here are a few highlights. Like most inventors, John Garand was somewhat eccentric. The Garand rifle was originally chambered for the .276 Pederson cartridge, but Doug MacArthur was Chief of Ordinance in the 30's and he decreed (wisely in my opinion) that all US rifle-caliber weapons be 30-06, reducing the M-1 Garand's capacity from ten to eight rounds. The M1903 sling was also compatible with the Garand, and the US ammo pouch accepted the 8-round Garand clips as easily as the 5-round M1903 clips. The M-1 Garand was adopted by the US Army in 1936 but produced in small numbers. In 1941 the Garand rifle beat the Johnson rifle and the M1903 as the USMC rifle. After the US entered the war, production could not keep up with demand, and an M-1 Garand was not in the hands of every American rifleman until 1943-44, although the M1903 continued for sniping and grenade-launching. The Germans had a semi-automatic rifle called the Gewehr 43 which was a good semi-automatic weapon, but simply was not as reliable and robust as the M1 (due to being made of crude parts and machining from mass supply shortages) nor was it as easly to produce or mantain in the field as the Garand rifle, so it was used mostly as a sniping rifle instead of a battle rifle, even then, it was very rare to encounter since it was never mass produced. The M-1 rifle fought again in Korea with the US and other armies, and it was replaced by the the M-14 in 1956. The US M-14 and the Italian BM59 are near-copies of the M-1 Garand in operation, but firing the smaller .308 Winchester cartridge.
How do you find the variance of a set of numbers?
1. Find the arithmetic mean (average)
2. Find the distance of each number from the mean (x-the mean=?)
3. Square each number
4. Add up all of the squares
5. Divide the final sum by the original mean
EX: Set: 1, 3, 4, 5, 7
1. 1 + 3 + 4 + 5 + 7 = 20 / 5 = 4
2. 1-4=-3 3-4=-1 4-4=0 5-4=17-4=3
3. 3*3=9 1*1=1 0*0=0 1*1=13*3=9
4. 9 + 1 + 0 + 1 + 9 = 20
5. 20 / 4 = 5
What is the probability that by guessing at random a person can guess the birth month of a stranger?
1 in 12
7/26
that it is a marble? 1. Unless I have lost my marbles.
What is the probability of a Fair six sided number cube rolling eight?
Assuming the numbers are 1-6, the probability is 0.
The odds are 50/50. A tossed coin does not have a memory.
What is the probabiltiy that a coin will land on heads in six tosses?
The probability that a coin will land on heads - at least once - in six tosses is 0.9844
What is probability and how does it helps?
Probability is the calculation of the extent to which something is probable (the likelihood of something happening or being the case). Probability helps staticians and other people to have a better understanding of what is to come in the future. Ex: If I have a bag of 10 colored marbles where five of them are blue and five of them are red, once I pull out a certain number of red and blue marbles I still know the ratio of red to blue marbles in the bag though it may have changed.
About 6
You want the area under the normal curve between z1 = (36-36.2)/3.7 and z2 = (37.5-36.2)/3.7
Using half tail tables (which give the probability (area under the normal curve) between the mean and z - the number of standard deviations from the mean of the value, negative just means it's to the left of the mean):
z1 = (36 - 36.2) / 3.7 ≈ -0.0541
z2 = (37.5 - 36.2) / 3.7 ≈0.3514
→ area between -0.0541 and 0.3515 standard deviations from the mean
= 0.0199 + 0.1368
= 0.1567
= 15.67 %
What is the probability of not getting a 2 or not getting a 5?
The answer depends on what the experiment is: drawing a card, rolling a die, with a spinner, ...
The answer depends on what the experiment is: drawing a card, rolling a die, with a spinner, ...
The answer depends on what the experiment is: drawing a card, rolling a die, with a spinner, ...
The answer depends on what the experiment is: drawing a card, rolling a die, with a spinner, ...
What is the intersection of the assessed probability and severity of hazards called in the crm?
It is the risk.
What kinds of things follow the normal distribution?