Probability

Suppose you could call it the Gaussian Distribution or the Laplace-Gauss (not to be confused with the Laplace distribution which takes an absolute difference from the mean rather than a squared error)... however the Brits had no one to name this distribution after (not the German and French names) and because it is the ubiquitous distribution they just called it... well the NORMAL!!

This is a law of addition probability which states that the probability of A or B equals the probability of A plus the probability of B minus the probability of A and B. Written in mathematical terms, the equation is: P(AorB) = P(A) + P(B) - P(AnB) where P(AnB) = 0 (since you can not pull out a green and black ball at the same time). Let P(A) = Probability of drawing the green ball & let P(B) = Probability of drawing the black ball. Total outcomes is 17. So, P(A) = 4/17 & P(B) = 6/17. Therefore P(green or black) = 4/17 + 6/17 = 10/17.

100*14/400 = 3.5%

I will deal with you later

Probability is defined as the number of ways an outcome can occur divided by the number of possible outcomes. For the coins, there are 4 outcomes (HH, HT, TH, TT). On the cube, there are 6 possible outcomes. The total number of outcomes is then 4*6 = 24. Since there is only 1 way to obtain HH, look at the cube outcomes. With the HH outcome, the cube would need to fall on a 4. So, there is only 1 way a HH4 can occur. Therefore the probability of getting 2 heads and a four is 1/24 or 0.04167.

5. Assuming the first four are all different suits, the 5th card must be a duplicate suit. If any prior to the 5th card is not a new suit, it is garunteed to be a duplicate of a prior suit.

An event that will definitely happen is an event with 100% probability.

Charts typically show and list the area to the left of the Z-Score value. To find the area to the right, just subtract the Z-Score value from 1; e.g. if the Z-Score value is .75 then take 1-.75 = .25.

Stratified Random Sampling. Google it. .

10*9*8=720

The sample space is 23 or 8; which can be listed out as: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. There are 2 of the 8 that have exactly 2 heads; so the probability of exactly two coins landing on heads is 2/8 or 1/4.

1) Simple random sampling

2) Systematic sampling

3) Stratified sampling

4) Cluster sampling

5) Probability proportional to size sampling

6) Matched random sampling

7) Quota sampling

8) Convenience sampling

9) Line-intercept sampling

10) Panel sampling

"The same as getting heads the first time, a one-in-two chance."

No It's Not! "One in two chance" means 50/50 chance. The odds of getting 2 in a row are not 50/50. I don't know what it is because that's how I arrived at this question. But if my friend were to bet me that he could get to heads in a row.... I'd take him up on that bet everytime, saying that he couldn't. He'd win a couple times here and there but in the end would lose money. It's not 50/50.

First suppose the two consecutive integers bex and x+1then according to the situationx+x+1=452x+1=452x=45-12x=44x=44/2x=22now our assumed consecutive integers were x and x+1 , Now put x in itx=45x+1=45+1 =46Thus the Solution set is {(45,46)}

The binomial distribution is defined by two parameters so there is not THE SINGLE parameter.

1. hello!!

The probability of drawing 3 hearts in a row in a 52 card deck is 13/52 * 12/51 * 11/50 = 1716/132600 = 0.01294, or approximately 1 chance in 77.

The probability of drawing a king is 4:52
The probability of drawing a diamond is 13:52 (or 1:4)
The probability of drawing a king (0.07692...) then replacing that king into the deck then drawing a diamond is 0.019230769.

If you leave the king out, the probability will be slightly greater (4/52) * (13/51)

Unless the king you left out of the deck was a king of diamonds, in which case, the probability would be (4/52) * (12/51)

We need to calculate two things:
How many possible possible series of 10 coin flips are there? As we flip 10 times and each time we can have either heads or tails we have 2 by the power of ten possibilities, or a total of 1024 unique possible series.

Now, how many of those series have exactly five heads and five tails? Lets assume we have ten "pre filipped" coins at hand - 5 tails and 5 heads. How many possible combinations are there. Well, if they were all different, you would have 10! (10 factorial = 10*9*8*7*6*5*4*3*2*1) possibilities.
How ever, the 5 heads are identical and so are the 5 tails, so if I interchange the locations of two coins that are both heads for example I still get the exact same series. There are 5! possible heads combinations, and 5! tails combinations.
Thus, the total number of unique combinations is 10!/(5!*5!) which happens to be 252.

So, out of 1024 possible series, 252 contain exactly 5 heads.
The probability thus is 252/1024=0.24609375 (roughly 25%)

the outcomes are 50:50.

1/3

Let me denote -A as the event that A does not happen. So we want Pr[-(A and B)]

Now, the event that neither A nor B occurs is the opposite of either A occurring, or B occurring or both occurring.

So Pr[-(A and B)] = 1 - Pr(A or B)= 1 - [Pr(A) + Pr(B) - Pr(A and B)] (since A+B is double counted)= 1 - (0.5 + 0.7 - 0.4)= 1 - 0.8= 0.2

This is a Binomial Probability Distribution; n=4, p=0.6. The probability of at least 1 passed is equal to the probability of 1-none passed; so x=0. The probability of x=0 (with n=4, p=0.6) is 0.0256. So, the probability of at least 1 passed is 1-0.0256 or 0.9744.

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