There are 52 cards in the deck.
The probability of drawing the ace of spades on the first draw is 1/52 .
Since you don't put the first card back, there are then 51 cards in the deck.
The probability of drawing the 4 of spades on the second draw is 1/51 .
The probability of both occuring is (1/52) x (1/51) = 1/2,652 = 0.037707 % (rounded)
There are 26 red cards and 13 spades in a standard deck of 52 cards. The probability of drawing a red card or a spade in one draw is, therefore, 39 in 52. If you draw twocards, and the first is not red or spade, then the probability on the second draw is 39 in 51, otherwise it is 38 in 51.Combining these two probabilities is easy. Just turn the problem around, and ask what is the probability of drawing two clubs? The answer is (13 in 52) times (12 in 51), which is 156 in 2652, or 1 in 17. Flip that answer over by subtracting it from 1, and you get a probability of drawing a red card or a spade in two draws of 16 in 17, or about 0.9412.
The answer will be 1.
As the number of times that the experiment is conducted increases, the experimental probability will near the theoretical probability - unless there is a problem with the theoretical model.
Divide
First tell me a problem and might understand
This problem is the type of the probability of A and the probability of B. These events are independent. P(A) and P(B) = P(A) * P(B). In this case these two probabilities are equal; the probability of a king is 4/52. So, the probability of draw king, replace and draw king is 4/52 * 4/52 = 0.00592.
You need to state a problem. If for example you ask what is the probability of drawing a heart or diamond in a single draw from a standard deck of 52 cards the answer would be .5
cluster sampling
There are 26 red cards and 13 spades in a standard deck of 52 cards. The probability of drawing a red card or a spade in one draw is, therefore, 39 in 52. If you draw twocards, and the first is not red or spade, then the probability on the second draw is 39 in 51, otherwise it is 38 in 51.Combining these two probabilities is easy. Just turn the problem around, and ask what is the probability of drawing two clubs? The answer is (13 in 52) times (12 in 51), which is 156 in 2652, or 1 in 17. Flip that answer over by subtracting it from 1, and you get a probability of drawing a red card or a spade in two draws of 16 in 17, or about 0.9412.
The answer depends on that the problem is!
1. Simple Random Sampling (SRS) - For SRS, every element has an equal probability of being chosen. In fact, any pair, triplet, and so on of elements have an equal chance of random selection. Sometimes, SRS can have problems because the randomness of the sample does not represent the population. For example, a SRS of one hundred people will likely produce about fifty men and fifty women, but it's also possible that there will only be ten men and ninety women selected due to natural sampling variation. 2. Systematic Sampling - For this type of sampling, every nth element is sampled. For example, if names were to be sampled through systematic sampling, every tenth name would be picked from the telephone book. However, this type of sampling may result in an unrepresentative sample of the population. 3. Stratified Sampling - When a population has certain categories, samples can be purposely collected from each strata (category). For example, there may be different strata for age groups if the person sampling is interested in variations between differences in age. One problem with stratified sampling is that it requires a more expensive cost than simple random sampling or systematic sampling. 4. Convenience Sampling - This type of sampling involves drawing the easiest samples to reach from the population. This may include surveying customers outside of a grocery store. Because the sample is limited to a certain time/day, it is unrepresentative of the entire population.
This is a sampling problem where the entire population is selected and there is no replacement. The chance of selecting all cities in alphabetical order is: (1/10)*(1/9)*(1/8)*(1/7)*(1/6)*(1/5)*(1/4)*(1/3)*(1/2) of 1/(10!) = 2.75 e-7. If we consider alphabetical order can be highest to lowest or lowest to highest, then the probability doubles or: 2/(10!)
The answer will be 1.
As the number of times that the experiment is conducted increases, the experimental probability will near the theoretical probability - unless there is a problem with the theoretical model.
Divide
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Scale drawing.