From a uniform distribution why is the total area under the curve equal to 1?
Because the area under the curve is a probability and probabilities range from 0.00 to 1.00 or could also be written as 0% to 100%
Assuming that you're only taking out one marble, then:
Your sample space --> 3 + 5 + 2 = 10
The probability of getting a blue marble on the first draw is 3/10 or 0.3
If you flipped a coin ten times what is the probability you would get five heads?
We need to calculate two things:
How many possible possible series of 10 coin flips are there? As we flip 10 times and each time we can have either heads or tails we have 2 by the power of ten possibilities, or a total of 1024 unique possible series.
Now, how many of those series have exactly five heads and five tails? Lets assume we have ten "pre filipped" coins at hand - 5 tails and 5 heads. How many possible combinations are there. Well, if they were all different, you would have 10! (10 factorial = 10*9*8*7*6*5*4*3*2*1) possibilities.
How ever, the 5 heads are identical and so are the 5 tails, so if I interchange the locations of two coins that are both heads for example I still get the exact same series. There are 5! possible heads combinations, and 5! tails combinations.
Thus, the total number of unique combinations is 10!/(5!*5!) which happens to be 252.
So, out of 1024 possible series, 252 contain exactly 5 heads.
The probability thus is 252/1024=0.24609375 (roughly 25%)
What is the probability of drawing 3 cards that are all queens from a deck of cards?
Aprox. 0.018%
There are 4 queens in a regular deck of 52 cards.
The probability of drawing a queen on the first draw is: P(Q1) = 4/52.
The probability of drawing a queen on the second draw given that the first card was
a queen is: P(Q2│Q1) = 3/51.
The probability of drawing a queen on the third draw given that the first two cards
were queens is: P(Q3│(Q2UQ1)) = 2/50.
The probability of drawing 3 queens on the first 3 cards drawn from a deck of cards
is: P(Q1UQ2UQ3) = (4/52)∙(3/51)∙(2/50) = 1.80995... x 10-4 ≈ 0.00018 ≈ 0.018%
If the two numbers are chosen randomly, they are independent so the fact that the value of the first is completely irrelevant. The probability that the second is more than 0.75 is the probability that a random number, between 0 and 1, is greater than 0.75 and that probability is 0.25.
The probability would be 1/32. Possible answers that fit the conditions would be .2,1 .2,.95 but .95,2 will not work. what you have to realize then is that .96-.2 has to be greater than .75 as stated. Because there is an infinite amount of numbers, or points on a number line, you must find length, or in this case, area. Create a graph, with an x-axis and y-axis. Now draw a square with side lengths 1. This area is equal to one, or 100 percent. By finding the area of numbers, or coordinates, that fit the condition, you can find the probability out of 100 percent that fits. Above, it says .96-.2 has to be greater than .75, so put into variables, that means y-x>.75 which simplifies to the inequality y>x+.75. Graph this as if it were y=x+.75 (or put it into a graphing calculator if you don't know how to do so). Now isolate the triangle that is formed by the line you drew and the square drawn earlier. this is an isosceles triangle, because the line has an elevation of 45 degrees. the side lengths are the same, and because there is .25 between .75 and 1, then both leg lengths of this triangle are .25. Now find the area of this triangle, and that is your answer
1/15 or about 0.07
sample space
Probability of dealt of a heart?
In a 52 card, 4 suit deck, the probability of selecting a heart is 13/52 or 1/4.
Will a coin land on tails more often if you flip it?
The chances are always the same:1/2. Sp it really depends on your luck, because if you flip a coin 100 times, it doesn't necessarily mean that tails will come up 50 times.
According to the Law of Large Numbers, as the number of times you flip a fair coin approaches infinity, the proportion of tails will approach 0.5 (as will that of heads).
What is the probability of what?
Guessing them all correctly?
Getting half of the correct?
Getting them all wrong?
PLEASE be specific with your questions if you want WikiAnswers to help.
Difference between weighted and numeric distribution?
numeric distribution is simply the number of items that have that thing divided by he total number of the sample, regardless of size or contribution.
weighted distribution is the number of items that have that thing multiplied by the weight of that thing divided by the total number of the sample.
example in sales
if a group of 100 retailers is the sample, 70 of them are keeping a certain product and the other 30 do not
then the numeric distribution of the product in this group is 70%
if all retailers have the same contribution to the sales to consumers, then the weighted distribution is 70% too
but if those 30 retailers (who do not keep have the product) contribution to the sales to consumers is 60%, then the weighted distribution of the 70 retailers having that product is only 40%
What is the normal distribution for clothing sizes?
You can find regulations about clothing sizes in the EN 13402 standard, and a series of physical measurements in the SIRI-dataset.
Reading the standard, I see that t-shirt sizes (men), for example, are mainly based on chest circumferences. Size 'M' is suitable for chest circumferences between 94 and 102 cm. Size S is 8 cm smaller, size L is 8 cm bigger, XL is 16 cm bigger and XS is 16 cm smaller than size M.
When I calculate the median and standard deviation of all the chest circumferences (adult males) I find in the SIRI-dataset, I find a median of 99.6 cm, and - surprise - a standard deviation of 8.4 cm. So, I tend to believe that clothing sizes follow, in some way, the normal distribution.
Size M refers to the median size, and the intervals between the size codes have about the same value as the standard deviation. So, size S is one standard deviation smaller than size M, and XL is two standard deviations bigger than size M.
If haven't checked other types of clothes and other physical sizes, so I cannot guarantee that my conclusion is correct for any type of garment.
If a standard number cube or die is rolled what is the probability that a 4 does not land face up?
If a standard number cube or die is rolled, the probability that a 4 does not land face up is five out of six, or (six minus one) out of six.
Set up the problem as: y=number of yellow marbles & 4y=number of blue marbles. 40=y+4y or 40=5y or y=8. Therefore number of blue marbles is 4*8 = 32.
67% as it's +/- one standard deviation from the mean
Two unbiased dice are thrown Find the probability that neither double nor a total of 10 will appear?
The sample space is 62 or 36. The easier way to do this problem is to find the probability of doubles or 10 and subtract from 1. The probability of a double is 6/36 (1,1; 2,2; 3,3; 4,4; 5,5; 6,6). There are 3 ways to get a 10; 4,6; 6,4; & 5,5. We already counted 5,5 in the double so there are 2/36 we add to the 6/36 to obtain 8/36. So, 1-8/36 = 28/36 or 7/9 or 0.7778 probability that neither double nor a total of 10 will appear.
14/27 ~ 0.519 ~ 51.9%
----------------------------------------------------------------------------------------------------
EXPLANATION
For bag 1: Probability of drawing a black ball; P(B1) =
6/9. A white one; P(W1) =3/9
For bag 2: P(B2) =
5/9. P(W2) =
4/9.
Probability of drawing a black ball from each bag: P(B1UB2) =
(6/9)∙(5/9) =
10/27
Probability of drawing a white ball from each bag: P(W1UW2) =
(3/9)∙(4/9) =
4/27
The probability of drawing 2 balls of the same color is the sum of the above
probabilities:
P(2 same color balls) =
P(B1UB2) + P(W1UW2) =
10/27 +4/27 =
14/27 =
0.5185...
P(2 same color balls) ~ 0.519 ~ 51.9%
What is the probability of rolling a 7 or 11 with two dice?
The probability of rolling a 7 with 2 dice is 6/36; probability of rolling an 11 is 2/36. Add the two together to find probability of rolling a 7 or 11 which is 8/36 or 2/9.
125.
Each job can be given to any of the 5 employees.
5*5*5 = 125.