Probability

Sample space, roll of 1 die, is: 1, 2, 3, 4, 5, 6. The numbers greater than 3 are: 4, 5, 6; which is 1/2 of the sample space. So, probability of rolling a number greater than 3 on one roll of a die is 1/2 or 0.5.

6*6=36

1&1; 1&2; 1&3; 1&4; 1&5; 1&6

2&1; 2&2; 2&3; 2&4; 2&5; 2&6

3&1; 3&2; 3&3; 3&4; 3&5; 3&6

4&1; 4&2; 4&3; 4&4; 4&5; 4&6

5&1; 5&2; 5&3; 5&4; 5&5; 5&6

6&1; 6&2; 6&3; 6&4; 6&5; 6&6

For adding up roll one and roll two, the outcomes are

1+1= 2

1+2=2+1= 3

1+3=2+2=3+1= 4

1+4=2+3=3+2=4+1= 5

1+5=2+4=3+3=4+2=5+1= 6

1+6=2+5=3+4=4+3=5+2=6+1= 7

2+6=3+5=4+4=5+3=6+2= 8

3+6=4+5=5+4=6+3= 9

4+6=5+5=6+4= 10

5+6=6+5= 11

6+6= 12

Odds of adding up to those possible outcomes.

2 = 1:36

3 = 2:36 = 1:18

4 = 3:36 = 1:12

5 = 4:36 = 1:9

6 = 5:36

7 = 6:36 = 1:6 You will total seven more times than others.

8 = 5:36

9 = 4:36 = 1:9

10= 3:36 = 1:12

11= 2:36 = 1:18

12= 1:36

In rolling two fair dice, the probability of obtaining a sum greater than 10 is 3 in 36 or 1 in 12, and the probability of obtaining a sum divisible by 6 is 6 in 36 or 1 in 6. Since the two probabilities come from the same population of 36 possible outcomes, the combined probability is the inclusive or of the two result sets, 8 in 36 or 2 in 9.

The results that match the sum>10 target are 5-6, 6-5, and 6-6.

The results that match the divisibility by 6 target are 1-5, 2-4, 3-3, 4-2, 5-1, and 6-6.

The results that match the combined target are 1-5, 2-4, 3-3, 4-2, 5-1, 5-6, 6-5, and 6-6.

There are six possible outcomes. Assuming the probability of each outcome is the same (dice has no defects), then you are likely to roll the number two, 100/6=50/3=16.67 times.

No standard deviation can not be bigger than maximum and minimum values.

18

False

False

The probability of throwing exactly 2 heads in three flips of a coin is 3 in 8, or 0.375.

There are 8 outcomes of flipping a coin 3 times, HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT. Of those outcomes, 3 contain two heads, so the answer is 3 in 8.

First you need to figure out the probability of it not landing on a 2 when you roll one die. Since there are six sides and two sides with 2 on it, the probability that one die will not land on 2 is 4/6.

Now you multiply 4/6 by 4/6 to find the correct odds as to both die not landing on 2.

4/6 * 4/6 = 16/36 = 4/9

The percentage is (4/9)*100% = (approx.) 44.4%

Prob(Sum is odd) = 1/2

Prob(Sum > 6) = 15/36 = 5/12

Prob(Sum is odd or >6) = 27/36 = 3/4

YES.

The standard deviation tells us nothing about the mean.

y = a * bxUsing "^" for powers. Any exponential formula can be written in one of the following two forms:

y = A e^Bx

y = C D^x

where "y" is the independent variable, "x" is the dependent variable, and "A", "B", "C", and "D" are arbitrary constants.

Considering the fact that a non-leap year having 28days in February will definitely have 4 Tuesdays , we have to find the probability that the 29th day of a leap year (2008) is also a Tuesday. This probability is 1/7 i.e 1 out the probable 7days in a week !

Well this said, this question is void in a sense knowing that we are in 2010 and that February 2008 had 5 Fridays and therefore the probability of 5 Tuesdays is ZERO as of 2008 !

Let z1 be the smaller z value and z2 be the larger value. Let N(z) be the cumulative normal distribution evaluated at some value z. The random variable is Z.

The probability that Z has a value from minus infinity to z2 is equal to N(z2). You can show this by drawing a bell shape curve, and shading in everything to the left of z2 as equal to the area under curve.

Similarly, the probability that Z has a value from minus infinity to z1 is N(z1).

The area under the bell curve (standard normal cumulative distribution) is N(z1) - N(z2). I can show this with a little example:

z1= -1 z2 = 2 Area = N(2) - N(-1) = 0.9773 - 0.1587 = 0.8186. I used Excel with the normsdist(z) function. The mean is zero and standard deviation is one with this function.

It means that the observations are all close to their mean value.

it is the test one tail

Ten of them.

48 marbles

Birthdays are not uniformly distributed over the year. But, if you do assume that they are, then:

ignoring leap years, it is approx 0.5982.

If you include leap years it is 0.5687.

If two dice (not dices!) are thrown then the probability that you get an even number on one die is 27/36 = 3/4

It means that it is less than the mean.

Use the entire excerpt of your printable table of random numbers, shown below, to estimate the answer to this question: What is the probability that a group of five random digits will contain at least three even digits? (Zero is considered an even digit.)

46370 55170 53480 49126 89212 75292 67291 88241 37808 38154

Express your answer as a fraction using the slash ( / ) bar. Reduce fractions to their lowest terms.

9/10