David has a stack of 30 cards, which are numbered 1 through 30. what is the exact probability of randomly selecting a card from
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Answer:
a) For this case we have that n = 30 and p =1/6
b) For this case we have that n = 10 and p =1/100
c) For this case we have that n = 20 and p =0.3
Step-by-step explanation:
Previous concepts
A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
The probability mass function for the Binomial distribution is given as:
Where (nCx) means combinatory and it's given by this formula:
Assuming the following question : "For each of the following binomial random variables specify n and p"
Solution to the problem
a. A fair die is rolled 30 times. X = number of times a 6 is rolled.
For this case we have that n = 30 and p =1/6
b. A company puts a game card in each box of cereal and 1/100 of them are winners. You buy ten boxes of cereal, and X = number of times you win.
For this case we have that n = 10 and p =1/100
c. Jack likes to play computer solitaire and wins about 30% of the time. X = number of games he wins out of his next 20 games.
For this case we have that n = 20 and p =0.3
There are two ways to do this but the way I prefer is to make one of the equations in terms of one variable and then 'plug this in' to the second equation. I will demonstrate
Look at equation 1,
this can quite easily be manipulated to show 
.
Then because there is a y in the second equation (and both equations are simultaneous) we can 'plug in' our new equation where y is in the second one
which can then be solved for x since there is only one variable 
and then with our x solution we can work out our y solution by using the equation we manipulated 
.
So the solution to these equations is x=-2 when y=6
Look at equation 1,
Then because there is a y in the second equation (and both equations are simultaneous) we can 'plug in' our new equation where y is in the second one
So the solution to these equations is x=-2 when y=6