Probability

You need to define one single variable, based on the six possible outcomes, such that the outcome of each trial is either a success or not.

Thus you could define X as "roll a 5" so that the probability of success is 1/6,

Or that X is "roll an even number", so the probability of success is 1/2 , or some other event.

The die need not be fair, but if it is loaded, the loading must not change. This can allow you to increase the range of probabilities of "success".

You then need to roll the die many times and record whether or not your chosen event occurred or not. The number of times the event occurred divided by the number of rolls will approximate a binomial distribution.

1/36

• queen card is 11
  • king card is 12

1/13 x 1/13 = .592%

Since there are 6 sides to the die, the probability of rolling a 5 on one roll is 1/6. Since each roll is an independent event the probability of the multiple results is the product of the probability of each result. So 2 consecutive 5's would occur with a probability of (1/6)(1/6) = 1/36

A public event is basically an event (party, gathering, meeting), where everyone can attend. For example: It could be a barbeque at a local Park. If it was a private event, that means that only certain people are allowed to attend. The best way to explain a private event would be a celebrities funeral. Only family members and close friends would be on the list to attend.

It is the process of deciding how large a sample is required so that the variable(s) of interest can be estimated to the desired degree of accuracy.

It is the process of deciding how large a sample is required so that the variable(s) of interest can be estimated to the desired degree of accuracy.

It is the process of deciding how large a sample is required so that the variable(s) of interest can be estimated to the desired degree of accuracy.

It is the process of deciding how large a sample is required so that the variable(s) of interest can be estimated to the desired degree of accuracy.

If you have 6 dice and you roll them one at a time (to explain this more easily but it works if you roll them simultaneously).

It doesn't matter what your first number is, so the odds of you getting that number is 1/1.

The second roll you have a 1/6 chance of getting the same number and this will be the same for the third, fourth, fifth and sixth dice.

Thus, you multiply each of these odds out.

1 x 1/6 x 1/6 x 1/6 x 1/6 x 1/6

Which equals a 1/7776 chance of rolling six of a kind in any one go.

If you have already rolled a six of a kind, it will still be a 1/7666 chance of getting it again because your previous actions do nothing to influence the chances of your future actions.

If you are wondering what the chances of rolling two 6 of a kinds over 2 goes you have to multiply the chances of getting a 6 of a kind by the number of times you want to get them (so (1/7776)2) which equals 1/60466176.

The formula used for working this out is (1/Ns)Nd-1

Where Ns is the number of sides on each die and Nd is the number of dice. It is minus 1 because it doesn't matter what is thrown on the first roll. If it matters what you roll (e.g. you have to throw a six every time), then on your first roll you have a 1/6 chance of rolling that number and so the -1 is not necessary in the formula.

Since each coin would have the outcome with Heads and Tails:

Then among the 32 coins, we can have the possible outcomes from no Heads, 1 Head, 2 Heads, ....... , 31 Heads, 32 Heads.

Therefore we would have 33 outcomes.

14.4%

The probability of a boy is approx 51.22%, which is very close to the international mean.

The answer depends on how many sides the spinners have.

1/2 * 1/6

The theoretical probability of HT or TH when two coins are tossed is 1/2 . (All possible outcomes are HH,TT,HT,TH). This means that when we run the experiment repeatedly we expect to get the desired result 1/2 of the time. Since you intend to toss the coins 40 times, 20 are expected.

A z-score of 70 would cover 35 standard deviations away from the mean. Note though, that a z-score of just 2.5 already covers 99% of the data. A z-score of 70 is incredibly high, and so is either a mistake, or will cover 100% of the data without fail. If a data point lies outside this, it is definitely an outlier and probably an error.

White Makrana marbles

mushroom

6.3

The fact that twins run in families is actually a myth. The odds of twins having twins is the same as a non-twin having twins. That is about 1 in 33.

No.

It is partially dependent on how many friends you have. With 6 friends, the chance is over 77% that two of them will have a birthday in the same month.

If you search "birthday problem" you will find many very detailed descriptions of how the formula works.

rare

answer = 1 - probability you do get two green*

* = binominal distribution f(k;n,p) = f(2;2,1/4)