There are 210 of them and they are:0123, 0124, 0125, 0126, 0127, 0128, 0129, 0134, 0135, 0136,
0137, 0138, 0139, 0145, 0146, 0147, 0148, 0149, 0156, 0157,
0158, 0159, 0167, 0168, 0169, 0178, 0179, 0189, 0234, 0235,
0236, 0237, 0238, 0239, 0245, 0246, 0247, 0248, 0249, 0256,
0257, 0258, 0259, 0267, 0268, 0269, 0278, 0279, 0289, 0345,
0346. 0347, 0348, 0349, 0356, 0357, 0358, 0359, 0367 ,0368,
0369, 0378, 0379, 0389, 0456, 0457, 0458, 0459, 0467, 0468,
0469, 0478, 0479, 0489, 0567, 0568, 0569, 0578, 0579, 0589,
0678, 0679, 0689, 0789, 1234, 1235, 1236, 1237, 1238, 1239,
1245, 1246, 1247, 1248, 1249, 1256, 1257, 1258, 1259, 1267,
1268, 1269, 1278, 1279, 1289, 1345, 1346, 1347, 1348, 1349,
1356, 1357, 1358, 1359, 1367, 1368, 1369, 1378, 1379, 1389,
1456, 1457, 1458, 1459, 1467, 1468, 1469, 1478, 1479, 1489,
1567, 1568, 1569, 1578, 1579, 1589, 1678, 1679, 1689, 1789,
2345, 2346, 2347, 2348, 2349, 2356, 2357, 2358, 2359, 2367,
2368, 2369, 2378, 2379, 2389, 2456, 2457, 2458, 2459, 2467,
2468, 2469, 2478, 2479, 2489, 2567, 2568, 2569, 2578, 2579,
2589, 2678, 2679, 2689, 2789, 3456, 3457, 3458, 3459, 3467,
3468, 3469, 3478, 3479, 3489, 3567, 3568, 3569, 3578, 3579,
3589, 3678, 3679, 3689, 3789, 4567, 4568, 4569, 4578, 4579,
4589, 4678, 4679, 4689, 4789, 5678 ,5679, 5689, 5789 and 6789.
Answer is Quota sampling.
Its one of the method of non-probability sampling.
The doctor in Macbeth suggests that the person with unnatural troubles should seek help from the divine (B) for a cure, rather than from sources such as witches, Hecate, or Banquo's ghost. This reflects the belief in the play that supernatural problems should be addressed through holy or divine intervention.
Luck can be influenced by various factors such as random chance, perception, and attitude. Some people might perceive themselves as consistently unlucky due to a negative mindset or focusing on negative experiences. It's important to remember that luck is often a combination of randomness and our interpretation of events.
You can use statistical tests appropriate for categorical data, such as chi-square tests or Fisher's exact test for associations between variables. For continuous data, you can use t-tests or non-parametric tests like Mann-Whitney U test or Kruskal-Wallis test. It's important to consider the limitations of quota sampling in interpreting the results.
The number of permutations of the letters SWIMMING is 8 factorial or 40,320. The number of distinct permutations, however, due to the duplication of the letters I and M is a factor of 4 less than that, or 10,080.
CRM stand for composite risk management. The CRM probability indicates whether or not a business transaction will actually take place.
because they can help you find your right answer
The question asks for the probability of an even card OR a red card. The term "OR" is key since this is not the same as the probability of drawing an even card and a red card, that is to say an even red card. GIven any two events, A and B P(A or B)=P(A)+P(B)-P(A and B) IF A and B are mutually exclusive, then P(A and B)=0 and this equation becomes P(A)+P(B) However, they are NOT in this case. So let A be the probability the card is even and B the probability it is red. P(A)=20/52 since J, K and Q are neither even nor odd (20=(52-12)/2)) P(B)=26/52 since half the cards are red. P(A and B) is the probability that a card is red AND even. We have 20 even cards, half of them are red and half are black so the odds are 10/52 of being red and even. P (A or B)=20/52+26/52-10/52=9/13
You can have:
1 row of 36
2 rows of 18
3 rows of 12
4 rows of 9
or 6 rows of 6, so in total there are 5 ways.
Let me give you a simple answer first and two deep answers next.
Simple answer is, assuming that all months have the same number of days (which is very approximately true), 1/12.
If you want a slightly deeper answer, then 31/365 or 31/366 is correct, depending upon whether the year is a normal year or a leap year.
The correct answer, which might be argued as "philosophical", is as follows:
The Gregorian calender, the current standard calender in most of the world, adds a 29th day to February, 97 years out of 400, a closer approximation than once every four years. This would be implemented by making a leap year every year divisible by 4 unless that year is divisible by 100. If it is divisible by 100 it would only be a leap year if that year was also divisible by 400.So, in the last millennium, 1600 and 2000 were leap years, but 1700, 1800, and 1900 were not. In this millennium, 2100, 2200, 2300, 2500, 2600, 2700, 2900 and 3000 will not be leap years, but 2400 and 2800 will be. The years that are divisible by 100 but not 400 are known as "exceptional common years". By this rule, the average number of days per year will be 365 + 1/4 - 1/100 + 1/400 = 365.2425.In 400 years, thereforethere are 303 normal years and97 leap yearsTotal number of days in 303 years110,595 days(normal years)Total number of days in 97 years35,502 days(leap years)Total number of days146,097 daysTotal number of August months400(in 400 years)Total number of days in August in12,400400 yearsProbability that a person is 0.084875 or 1 in11.78202born in August (=12,400 / 146,097) or 31 in365.24251 / 12 =0.08333331 / 365 =0.08493231 / 366 =0.084699
There are assumptions in the above answer as well. These are
1. That all days in a year have the equal probability (which is not strictly true, as records of births will be required for verification)
2. The Gregorian calender has been in effect since 1700s. The probability as above will only work for years after the introduction of Gregorian calender.
T-distributions tend to be flatter and more spread out than normal distributions due to their heavier tails. Unlike the normal distribution, which has thin tails, t-distributions account for uncertainty in sample variance estimation, making them more robust for smaller sample sizes. The additional variability inherent in t-distributions arises from the incorporation of the sample size through the degrees of freedom parameter. As the degrees of freedom decrease, the t-distribution becomes more spread out and flatter, reflecting increased uncertainty and variability in the estimates. This property makes t-distributions well-suited for inferential statistics, particularly when dealing with small sample sizes.
sample space
There are a number of retailers from where a rolling tote bag can be purchased. These retailers include The Find, Smart Crafts, Amazon and Marks and Spencer.
The probability is
(5 times the number of 6s on the spinner/6 timesthe total number of different positions on the spinner)
Assuming the die to be a cube, with each face having a number 1,2,3,4,5,6.
Since there are 6 digits in total and only three are selected .
The P(4,5,6) = 3 / 6 = 1/2 = 0.5
The probability is 0.
P(female offspring) = 0.5 = 1/2
Reason,
In the human genetic system, a male has 'XY' chromosomes
A female has 'XX' chromsomes. I will designate the femle chromosomes as X' & X" to distinguish .
Hence Male is XY
Female is X'X"
They can match in the following manner.
XX' = female
XX" = female
YX' = male
YX" male.
So you can see that there are two female outcomes and two male outcomes, out of a total of four(4) possible outcomes.
So 2/4 = 1/2 = 0.5 in both cases.
Hence P( female outcome) = 0.5
NB In reality, with a deeper understaning of statisitical and genetic analysis, slightly more females are born that males. Also males are slightly weaker at birth, there by making males less likely to survive at birth. This difference is only about M:F :: 49% : 51% , so equality of numbers remains roughly equal.
Q9. How many different ways can 4 tickets be selected from 50 tickets if each ticket
wins a different prize.
Probability of drawing a black 7 from a standard 52-card deck is 2/52 or 1/26.
85, because, in this case, you are most likely to choose 1 card of every rank.