What is the PROBABILITY of throwing 2 sixes with two dice?
This is an example of an INDEPENDENT event:
The probability of throwing a double six with two dice is the result of throwing six with the first die and six with the second die. The total possibilities are, one from six outcomes for the first event and one from six outcomes for the second, Therefore (1/6) * (1/6) = 1/36th or 2.77%.
The two events are independent, since whatever happens to the first die cannot affect the throw of the second, the probabilities are therefore multiplied, and remain 1/36th.
What is the sample space of tossing 4 coins?
If the order of the outcomes matters, then
TTTT,
TTTH, TTHT, THTT, HTTT,
TTHH, THTH, THHT, HTTH, HTHT, HHTT,
THHH, HTHH, HHTH, HHHT,
HHHH.
If the order does not matter, then
TTTT, TTTH, TTHH, THHH AND HHHH
15/27. Simply take the probability of drawing a blue marble.
How do you solve problems on probability?
There are many different problems and different ways for solving them.
Here number of sample points that is total possible outcomes would be 52C3.
Event A = Three cards draw are hearts.
So here we can select 3 hearts out of 13 hearts in 13C3 ways...
Therefore the required probability would be
=(13C3)/(52C3)
=286/22100
=0.01294
Kunal K.
Try calling it, "14 choose 10". Set up the equation: n!/r!(n-r)!
n=14, r=10 So: (14!)/[10!(14-10)!]
Final answer: 1001 different ways. The number is so large it is simply counter intuitive, isn't it?....but it's true!
The probability is (51.5-51.25)/(52-50) = 0.25/2 = 0.125
Probability of two tetrahedral dice rolled the sum on the upward faces is four?
When a tetrahedral die is rolled, it will rest with three faces upwards. If the die is numberd from 1 to 4. therefore the sum of the upward facing numbers on 1 die is at least 6 and so for two dice, the minimum is 12. That being the case, the probability is 0.
In a normal distribution with mean 80 and variance 144 squared what is the probability above 100?
The probability is 0.4448, approx.
In an ordinary 52 card deck, there are 4 three's, and 13 clubs, one of which is the 3 of clubs, so there are 16 cards that are a three or a club. The probability of drawing a three or a club is 16 out of 52, or 4 out or 13, or 0.308.
0
No blue marbles in the bag.
If you roll a die 60 times about how many times would you expect to get 1?
the probability 6o =1/6 = 60%
If 4 coins are tossed what is the probability of getting 3 heads?
C1, C2, C3, C4
The number of combinations is 4 base 2, or 16.
There are 4 possible scenarios where you get 3 heads.
So the odds are 4:16, or 1:4.
If five dice are rolled how many possible outcomes?
Depends on what detail of outcome you are interested in. If you are only interested in the total, you can get all numbers between 5 (5 ones) and 30 (5 sixes) - a total of 26 different results.
Probability of having 3 girls out of 4 children?
There is no simple answer to the question because the children's genders are not independent events. They depend on the parents' ages and their genes.
However, if you assume that they are independent events then, given that the probability of a boy is approx 0.52, the probability of 3 girls out of 4 children is 0.2331
If a couple has had 3 boys and wants to have a girl what is their probability of having a girl?
These events are independent; so the probability of a girl is 0.5.
The probability is approx 0.81
12 white marbles from (7+3+12) = 22 marbles
Chance of a white marble on first pick = 12/22 = 6/11.
Chance of a white marble on second and third picks are the same, as the marble is replaced.
So, the chance of a white marble three times is 6/11 * 6/11 * 6/11 = 216/1331 = approximately 16.23%
false
How do calculate the probability of at most?
For a discrete variable, you add together the probabilities of all values of the random variable less than or equal to the specified number.
For a continuous variable it the integral of the probability distribution function up to the specified value.
Often these values may be calculated or tabulated as cumulative probability distributions.
That defines the odds in favour of the event.