Assume that a procedure yields a binomial distribution with a trial repeated n times. Use a binomial probabilities table to find the probability of x successes given the probability p of the success on a given trial N=9 X=0 p=0.30
the equation you need is
P(X=r)=C(n,r)*p^r*(1-p)^(n-r)
C(n,r)=n!/((n-r)!r!)
and n!=n(n-1)(n-2)....(2)1=factorial n
You have everything you need to calculate the required probability.
It turns out that C(n,0)=1 for any positive value of n.
The answer is less than 5%. You can post your answer for a check if you wish.
To find the probability of x successes using a binomial probabilities table, you will first need to locate the row corresponding to the number of trials (n) and the column corresponding to the probability of success (p).
In this case, you are given that the number of trials is N = 9, the number of successes is X = 0, and the probability of success on a given trial is p = 0.30.
1. Begin by finding the row that corresponds to the number of trials (n = 9). Locate the row labeled "n = 9" in the binomial probabilities table.
2. Once you have found the row for n = 9, search along that row to find the column that matches the given probability of success (p = 0.30). Look for the closest value in the table. In this case, you might find a value near p = 0.3, such as p = 0.2995.
3. After finding the row and column that correspond to your given values of n and p, locate the intersecting cell in the table. The value in this cell represents the probability of obtaining exactly x successes (X = 0) in the given number of trials (N = 9).
So, using the binomial probabilities table, you would find the probability of X = 0 successes given N = 9 trials and p = 0.30 by locating the cell where row "n = 9" intersects with column "p = 0.2995".