Probability

The probability of an event is the chances it will happen divided by all the possible outcomes.

To solve this problem you: 1) Find all factors of 24 (there are 6 factors- 1,24,6,4,2,12) 2) You elemenate all numbers that are not on a six sided die (24,12) 3) Now use probability. The die is numbered 1-6 and you have numbers 1,2,4 and 6 which makes 4 out of 6. 4) This last step I can't show you, so I'll have to explain. You set up a proportion box. Write the 4 over the 6 on one side and next to the 6 write the 96. Over the 96 write N, which will represent the answer to your equation. Then multiply 4 times 96 and divide the product by 6. If you followed these steps you should get an answer of 64.

Yes there are 52cards in a deck without joker - its all depends on game or country which you play

vijay

yes

120

The probability of getting a head is 1/2 and if you toss it 4 times, the probability of 4 heads is (1/2)^4=1/16 since the tosses are independent events.

Let us assume you can repeat numbers. That is to say, we allow 101 and are not bothered by the repeat 1, or 111 for that matter. Let's do it first so that the order matters, that is to say 123 is not the same as 321. The possible numbers for the first digit are 0,1,2,...9 so there are 10 of them, Same for the second. Now for the digit "in front", for example in 123 the digit 1, we must decide if we can allow a 0. To make things easy let's say yes for now. Then there are 10 choices for each digits and 10^3 total choices=1000 Now, lets narrow things down just in case we have some restrictions. If exclude 0 from the digit in front, then there are only 9 choices for that place and we have 900 possibilities. Now you said combinations and that usually means that order does not matter. A permutation is where order matters. So if order does not matter 123 is the same as 321 or 213. This means we over counted and now must divided. Each group of 3 numbers such as 123 can be arranged in 6 different ways, which is 3! for example 123 132 213 232 312 321 We we divided by 3! Assuming the leading 0 was allowed we have 1000/3! combinations, if not we have 900/3! combinations.

At least 2 sixes mean 2,3,4, or 5 sixes. This is the sum of the probability of 2 sixes + probability of 3 sixes etc. It is easier to look at the probability of 0 or 1 six and subtract that the sum of those two from 1. P( one six)=1/6 P(number other than six)=5/6 P( one six in 5 tosses)=(1/6)(5/6)^4=625/7776 P (no six in 5 tosses) the probability of not getting a 6 of the first toss is 5/6 Now if we do not get one on the first toss, we have 5/6 chance of not getting it again on the second, and the third etc. (5/6)^5=3125/7776 If we add these we have 3750/7756 which is the odds of exactly 0 or 1 six. 1-(3750/77756)=2003/3878 .516503 is the probability of at least 2 sixes in 5 tosses.

its about 50/50, but mostly depending on the skin to of the grand parents

Any event where you want the event to occur randomly on one of the the 7 days of the week.

There are 6 outcomes with one roll and 6^2 with two and 6^3 with three. So the answer is 216.

The probability is .5 since half the numbers are even and half are odd.

The probability is 90/216 = 5/12

Probability of an event A is P(A) is the (chance of A)divided by (the total chances) and odd is (Chances for: Chances against) For example the probability of picking a 6 in a deck of cards is 4/52 and the odds are 4:52-4=4:48 or 1:12

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 5-4 5-6

6-1 6-2 6-3 6-4 6-5 6-6

So there ARE 36 possible outcomes, you see.

Answer BY: Magda Krysnki (grade sevener) :P

Given "n" random variables, normally distributed, and the squared values of these RV are summed, the resultant random variable is chi-squared distributed, with degrees of freedom, k = n-1. As k goes to infinity, the resulant RV becomes normally distributed.

See link.

try http://www.vedicmaths.webs.com

It means there is virtually no chance an event will happen.

There are 2 possibilities for each toss. Since the three tosses are independent (one trial does not affect the outcome of the other trials), there are 2 * 2 * 2 = 8 total possible outcomes.

The outcomes are:

HHH

HHT

HTH

HTT

THH

THT

TTH

TTT

Dependent events.

1/3

Planning your route is the best way to control events that occur outside your vehicle.

There are 6*6*2 = 72 possible outcomes.

Let M be the mean.

P( X > 5.52) =0.1271

Standardize X by subtracting the mean and dividing by the standard deviation.

z = (52.2-M) /4

From the normal distribution tables, P( z > 1.14 )=0.1271

Therefore, (52.2-M) / 4 = 1.14

Solve for M, the mean.

52.2-M = 4(1.14) = 4.56

M=52.2-4.56 =47.64

P(H and H and H and H)= 1/2 X 1/2 X 1/2 X 1/2 = 1/16