Probability

drawing a club

total club =13

P(Club)= 13/52

drawing a 7

total 7 = 4

P(7) = 4/52

P(a club and a 7) = P(Club) + P(7) as events are mutualy exculisve

P(a club and a 7) = 13/52+4/52

P(a club and a 7) = 17/52

Since there are 6 numbers on a die (1-6), then the probability of rolling a 5 would be 1 out of 6.

The probability of having a girl versus a boy is 1/2 because there is two things you have a chance of getting and you can only get one or the other.

judgemental is Pudi lan

Haphazard sample

12/20, 3/5, or 0.6

50%

1/2

It can be.

Coin tosses are independent events. The probability of a head remains 1/2

Average number of arriving customers in 4 minute interval is 2.8 and poison distributed what is the probability that exactly six customer will arrive in a 4 minute interval?

You can roll (3,6), (4,5), (5,4), or (6,3). So that's four possibilities out of 36. So 4/36 or 1/9.

Every time you flip a coin it has a 50% chance of heads and a 50% chance of tails. Flipping a coin multiple times does not change that. Therefore the answer is 50%

Five numbers can be arranged in 5! = 120 ways.

The first roll doesn't matter for probability, it just sets the number to be rolled by the other two. So:

P(rolling the same number three times) = P(rolling a particular number)2 = (1/6)2 = 1/36

The discrete random variable almost always arises in connection with counting.

The probability of the event occurring.

If they're disjoint events:

P(A and B) = P(A) + P(B)

Generally:

P(A and B) = P(A) + P(B) - P(A|B)

18

The probability is the number of girls divided by the number of students, so 12/22, or 6/11

Suppose the probability density function is f(x), defined over a domain D

Then the mean is E(X) = x*f(x) integrated with respect to x over D.

Calculate E(X2) = x2*f(x) integrated with respect to x over D.

Then Variance(X) = E(X2) - [E(X)]2

and Standard Deviation = sqrt(Variance).

A dependent event. Or rather, a dependent event is one whose probability of occurrence is affected by previous events. For instance, drawing a card from a deck is affected by previous draws, if there's no replacement.

{hhhhhh, hhhhht, hhhhth, hhhhtt, hhhthh, hhhtht, hhhtth, hhhttt, hhthhh, hhthht, hhthth, hhthtt, hhtthh, hhttht, hhttth, hhtttt, hthhhh, hthhht, hthhth, hthhtt, hththh, hththt, hthtth, hthttt, htthhh, htthht, htthth, htthtt, httthh, htttht, htttth, httttt, thhhhh, thhhht, thhhth, thhhtt, thhthh, thhtht, thhtth, thhttt, ththhh, ththht, ththth, ththtt, thtthh, thttht, thttth, thtttt, tthhhh, tthhht, tthhth, tthhtt, tththh, tththt, tthtth, tthttt, ttthhh, ttthht, ttthth, ttthtt, tttthh, ttttht, ttttth, tttttt}

Edit: It removed all my capitalization, but everything should be capitalized.

This problem doesn't lead to a whole number solution. Are you sure you copied it correctly? Maybe you meant the ratio is 4 to 3. In that case:

red/blue = 4/3

blue/red = 3/4

blue = (3 * red) / 4 = (3 * 16) / 4 = 12 blue marbles

Every roll of a fair die has a one in two chance of being even or odd.