Probability

Prob(at least 4 heads) = Prob(4 heads) + Prob(5 heads) = 5/32 + 1/32 = 6/32 = 3/16

1 in 13 - one out of every 13 cards is a queen.

Alpha or significance level is usually 0.1 or 0.05 for a 2-tailed test.

You can use a normal distribution to approximate a binomial distribution if conditions are met such as n*p and n*q is > or = to 5 & n >30.

The probability of event A occurring given event B has occurred is an example of conditional probability.

13 cards with hearts on them
A=1
2=2 etc.

Snowball sampling can be used in a number of circumstances.

Sometimes the researcher may be able to identify only a small number of individuals in the target group but these people are likely to known others in the target group, and people that they know will also know more such people-and so on. By including those few initial people in her sample, and asking them to recruit others, the researcher may be able to achieve a useable sample size.

Then, sometimes, a researcher is actually more interested in the connections between people, or in other words, social networks. In this case, snowball sampling can be used to identify those connections.

Please see the link.

In order to determine the probability of drawing 2 hearts and then a spade, in that order, from a deck of 52 cards, start by considering the first card.

The probability of drawing a heart is 1 in 4.

Since you have now reduced the number of hearts and the number of cards in the deck by one, the probability of drawing another heart is 4 in 17.

Since you have further reduced the number of cards by one, the probability of drawing a spade is 13 in 50.

Multiply these probabilities together, (1/4) (4/17) (13/50), and you get about 0.0153, or about 153 in 10000.

probability

90.47619%

Close but no cigar. I presume u r talking about rolling a pair of dice. There are 6 possible outcomes 4 each die, all equally likely, for a total of 36 equally likely possible outcomes. There is only 1 with a sum of 2, and 2 with a sum of 3, for a total of 3 outcomes with a sum of 3 or less. That's 3 out of 36, or 1/12. The other 11/12 represent a sum of more than 3, and that's 91.6666666...%

There are 11!/(4!*4!*2!) = 34,650 ways.

5, 3, 5, and 7 make a sum of 20. So 7/20 would be the probability of find a yellow one. Then, subtract 7 from 20 to find the probably of not finding a yellow one, which is 13/20, or 65%

There are two outcomes for each coin and three coins; 2 x 2 x 2 = 23 = 8 outcomes.

Percentage

Yes. the conditional expectation of X given Y is simply the expectation of X if X and Y are uncorrelated. This is a consequence of one of the properties of conditional expectation.

Coins cannot be "calculated".

There being 365 days in a year and 50 being less than 365 therefore 2 even far less than that the chances are virtually 0.

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Assume that all 366 days (including leap day) are equally likely to be a person's birthday. The probability that none of them share a birthday is 1*P(second person selected doesn't share a birthday with first person selected)*P(third person selected doesn't share a birthday with first or second person selected)*...*P(fiftieth person selected doesn't share a birthday with the first, second, third,...,forty-ninth person selected).

P(second person selected doesn't share a birthday with first person selected) = 365/366

P(third person selected doesn't share a birthday with first or second person selected)=364/366

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P(fiftieth person selected doesn't share a birthday with the first, second, third,...,forty-ninth person selected)=317/366

P(none share a birthday)=(365/366)*(364/366)*...*(317/366), which is approximately .0299.

P(at least two share a birthday) = 1-(365/366)*(364/366)*...*(317/366)=1-.0299=.9701 = 97.01%.

There are 23 possible outcomes this is found by there being 2 possibilities ( head or tails) that is done three times, that is 2X2X2= 8 possible combinations for the outcome. This is if order matters, that is h h t is different to t h h

5 to 1

well there are 52 cards in a pack and there are 4 different suits so its 13/52 which simplifies to 1/4

Yes, the mean (and median and mode) is the 50th percentile of any normal distribution.

...normally distributed.

1/12

P(8) AND P(8) = 4/52 * 3/51 = 12/2652 = 0.00452.

The de Moivre-Laplace theorem. Please see the link.