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Probability
The probability of a certain event is a number expressing the likelihood that a specific event will occur, expressed as the ratio of the number of actual occurrences to the number of possible occurrences. In mathematics, it is a measure of how often an event will happen and is the basis of statistics.
14,643 Questions
Derive the moment generating function of the poisson distribution?
The probability mass function (pmf, you should know this) of the Poisson distribution is
p(x)=((e-λ)*λx)/(x!), where x= 0, 1, ........
Then you take the expected value of exp(tx), you should always keep in mind to find the moment generating function (mgf) you must always do
(etx)*p(x), where t is a random variable
Therefore,
(etx)*((e-λ*λx)/(x!))
(e-λ)*sum[(e-λ*λx)/(x!)]
Thee-λ is only a constant; thus, it can be pulled out of the sums.
Continuing,
(e-λ)*sum[(λ*et)x)/x!]
Let y=λ*et
(e-λ)*sum[(y)x/x!]
By Macalurins series, the sum[(yx)/x! ]= ey
Soonwards
(ey)*(e-λ)
Lets return the y by λ*et
If you roll a number cube twice what is the probability of rolling a 6 and a 2?
Assuming that a dice roll is purely random, there is a 1 in 6 probability of landing on any number. Since the second roll depends on the probability of the first, they factor together:
First Roll: 1/6
Second Roll: (1/6)(1/6) = 1/36
If you were rolling both dice at once, however, the math would be completely different.
Probability of getting an even product when rolled 2 dice?
I'm assuming you multiply the numbers rolled. Thus the rules of even and odd numbers come into play. The only way to get an odd product is to roll two odd numbers. The probability desired is 1 minus the probability of rolling two odd numbers.
The odds of rolling an odd number is 1/2. Since the rolls are independent events, the odds of rolling two odd numbers is (1/2)2 = 1/4
So the probability of getting an even product is:
1 - 1/4 = 3/4
4:
Because there will be at least 1 of two kinds of marbles, and 2 of the other.
Ex: 1 black, 1 red, and 2 whites.
The Z test.
1/6
I'm assuming "genius" means an unusual value (more than two standard deviations) in the right tail. This is (100 - 95.45)/2 = 2.275 percent, approximately. So:
6,575,000,000 * 0.02275 = 149,581,250 geniuses ... not that special, huh?
What is the probability of rolling the same number 4 times?
The first number doesn't matter, just that the last three are the same as the first.
P(rolling the same as the number before) = 1/6
P(rolling the same number as the first three times = (1/6)3 = 1/216
9 In how many different ways can you arrange 7 books on a shelf?
Consider one placement at a time. The first book on the shelf could be any of seven. The second could be any of the remaining six. This continues with the rest. The total combinations are:
7! = 7*6*5*4*3*2*1 = 5040
1. It follows instruction A
2. It follows instruction B
3. It follows both instruction A and B
4. It follows neither instruction
If yellow equals one fourth what would the probability of spinning a yellow 50 times?
1/4 raised to the power of 50; e.g. .2550.
If a hexahedral die is rolled two times what is the probability of not rolling a five both times?
25/36
Discrete and Continuous data is used in mathematics?
Yes it is. Discrete data is something that's set. Like say you were making a line graph about renting bikes. You can only rent whole bikes there is nothing in between. You shouldn't connect points on a line graph with discrete data but some cases can be argued.
Continuous data is usually a measurement that could change like time.
What is the probability that a family of four children has at least 2 girls?
Going by the assumption that the probability of a child being a girl is 1/2, and that the events are independent, then the probabilities of different numbers of girls are as follows:
P(0 girls) = 1/2 ^ 4 = 1/16
P(1 girl) = 4 * (1/2) * (1/2 ^ 3) = 1/4
P(2 girls) = 6 * (1/2 ^ 2) * (1/2 ^ 2) = 3/8
P(3 girls) = 4 * (1/2 ^ 3) * (1/2) = 1/4
P(4 girls) = 1/2 ^ 4 = 1/16
Therefore the probability of at least 2 girls = 3/8 + 1/4 + 1/16 = 11/16
What is the probability of at least one birthday match among a group of 41 people?
The probability of at least 1 match is equivalent to 1 minus the probability of there being no matches.
The first person's birthday can fall on any day without a match, so the probability of no matches in a group of 1 is 365/365 = 1.
The second person's birthday must also fall on a free day, the probability of which is 364/365
The probability of the third person also falling on a free day is 363/365, which we must multiply by the probability of the second person's birthday being free as this must also happen. So for a group of 3 the probability of no clashes is (363*364)/(365*365).
Continuing this way, the probability of no matches in a group of 41 is (365*364*363*...326*325)/36541
This can also be written 365!/(324!*36541)
Which comes to 0.09685...
Therefore the probability of at least one match is 1 - 0.09685 = 0.9032
So the probability of at least one match is roughly 90%
What is the probability of flipping a coin 7 times and receiving all heads?
The probability of each coin coming up heads is 1/2 .
The probability of all 7 coming up heads is (1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2) = 1/128
That's 0.78 percent (rounded)
(Dr. Atticus was right.)
The expected number is 12/7 = 1.29 approx.
What is the probability of getting exactly 5 boys and 5 girls if you have 10 kids?
Here the sample space is(s)=10, =>mod(s)=10
a=be the event of getting exactly 5 boys =>mod(a)=5
b=be the event of getting exactly 5 girls =>mod(b)=5
thus,
p(a)=mod(a)/mod(s)=5/10=1/2
p(b)=mod(b)/mod(s)=5/10=1/2
p(5 boys and 5 girls)=p(a)*p(b)=1/2*1/2=1/4
Bala has 5, Ramu has 7
