Probability

Births are not distributed randomly across the year: there are seasonal variations as well as differences between days of the week. Where maternity staff like to celebrate New Year's Eve, there may be fewer people on duty on January 1 and so there may well be fewer births. Unfortunately, comprehensive information on such variations is not available.

If you assume that each day of the year is equally likely, the probability is 1/365. Adjusting for leap years changes that to 1/365.25. Further adjusting for Centuries gives 1/365.2425

Prob(The one block is blue) = number of blue blocks/total number of blocks = 8/25 = 32%

An F-statistic is a measure that is calculated from a sample. It is a ratio of two lots of sums of squares of Normal variates. The sampling distribution of this ratio follows the F distribution.

The F-statistic is used to test whether the variances of two samples, or a sample and population, are the same. It is also used in the analysis of variance (ANOVA) to determine what proportion of the variance can be "explained" by regression.

Since each digit has 4 possibilities to be arranged, and 2 of the digits have the same value of 3, the number of arrangements would be 12.

4!/2! = 4 x 3 x 2!/2! = 12

They go to a far country where they go

The probability that the birthdays of five persons chosen at random will fall in twelve different calender months is zero.

You would need at least twelve persons to have a non zero probability.

4

The main advantage is that the sample is representative of the population and the mean of the sample is an unbiased estimate of the population mean. Also, characteristics of other statistics based on the sample are well understood.

However, sometimes it may not be possible to gather valid information from a sampling unit and then the sample is no longer random. This can be either because the sampling unit cannot be located or has been compromised by external factors. This can be particularly serious if the "missing" units share a common characteristic.

Also, simple random samples may not include any units representing characteristics that are rare in the population - but important in the context of the experiment.

Independent. The chance the die are either a total of 11 or one will be a 4 are not related.

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There are only two ways the sum of the resulting numbers of two dice thrown give 11: (5,6) and (6,5).

The probability for event A is: P(A) = 2(1/6)(1/6) = 1/18 = 0.05555... ≈ 5.56%

Given event A, there is no possibility that one of the two dice show a 4.

The conditional probability is, P(B|A) = 0.

So, event B is dependent of event A.

1

Classical approach has possible outcomes which are known with certainity ie sampling distribution is known.

Relative approach is an approach in which probability values are based on historical interest.

z(.97) = 1.8808

Since z= (x-Mu)/s

x=z*s + Mu

x=1.8808*0.09g + 8.38g ~= 8.549 g

Multiplication Principle of Counting

True.

1/9 ways. total sample space for the experiment is 36(because you have two dice). and you can get 5 in 4 ways that is (1,4),(4,1),(2,3),(3,2). hence 4/36= 1/9.

The area under the standard normal curve is 1.

The sample space of a standard six sided die is [1,2,3,4,5,6].

1.14

Less. The more times the coin is tossed, the more likely it will reflect the actual odds of .5 heads and .5 tails.

5/20, 25%, or .25

P(a^~a)==1 P(a&~a)==0

the line above is shorthand notation for an event that has a probability of 1, followed by an event that has a probability of 0. P(event) is an easy way to say the probability of "event". The "^" means "OR", the "~" means "NOT", and the "&" as you are probably familiar means "AND". So puting it together, "P(a^~a)" means the probability that an event "a" occurs OR that event "a" does not occur. So take an event "a", any event, like drawing an 8 of clubs out of a deck of cards. If you draw a random card out of a deck of cards, the probability that you will draw an 8 of clubs OR that you will not draw an 8 of clubs is 1. That means 100%. So when you draw a card you will either draw the 8 of clubs or not draw the 8 of clubs. It seems like an obvious statement to make but its a proof that becomes very important in proving less obvious theories. likewise, the second statement was "P(a&~a)", so the probability that event "a" occurs AND event "a" doesnt occur. Since the event has to either occur or not, it cant occur AND not occur, so the probability is 0.

the probability of getting one head and one tail on three flips of a coin is 1/9

Yes, it certainly can if there is only one possible outcome. For instance, the probability of drawing a red ball from a bag containing nothing but red balls is equal to one.

mean = 5, variance = 5

Since there are 6 numbers (1-6) on each die, and you're rolling it 5 times, the total number of possibilities is 65. You now have to find out, out of the total, how many possibilities fill your requirements. For 5 rolls of one the chance is 1/65. There are 5 possible sets of rolls where you'll get 4 1s (4 1s and a 2, 3, 4, 5, or 6). Lastly there are 25 possible sets of rolls where you'll get 3 1s (3 1s and a 2 and 2 or a 2 and 3 or a 2 and 4 etc). so adding up all the possibilities that is 31/65. That is a percentage of .4% or 4 out of 1000 rolls of all the dice you will get 3, 4, or 5 1s. Sounds like a yahtzee question.