for n repeated independent trials, with constant probability of success p for all trials, find the probability of exactly x succes n=5,p=1/3, x=4
If there are only two possible outcomes from each trial then it's a binomial and by using the formula nCx p^x(1-p)^(n-x) we get; 5C4(1/3)^4(2/3)^1=0.04115
for n repeated independent trials, with constant probability of success p for all trials, find the probability of exactly x succes n=12,p=0.4, x=6
To find the probability of exactly x successes in n repeated independent trials, with a constant probability of success p for all trials, we can use the binomial probability formula:
P(x) = C(n, x) * p^x * (1-p)^(n-x)
Where:
- P(x) represents the probability of getting exactly x successes.
- C(n, x) represents the number of combinations of n items taken x at a time, which can be calculated as C(n, x) = n! / (x!(n-x)!)
- p is the probability of success on each trial.
- n is the number of trials.
- x is the desired number of successes.
In this case:
- n = 5 (as given in the question)
- p = 1/3 (as given in the question)
- x = 4 (as given in the question)
Now, let's calculate the probability using the provided values:
P(4) = C(5, 4) * (1/3)^4 * (2/3)^(5-4)
First, let's calculate C(5, 4):
C(5, 4) = 5! / (4!(5-4)!) = 5! / (4! * 1!) = 5
Now, substitute the values into the formula:
P(4) = 5 * (1/3)^4 * (2/3)^1
= 5 * (1/81) * (2/3)
= 10/243
Therefore, the probability of getting exactly 4 successes in 5 repeated independent trials with a constant probability of success of 1/3 is 10/243.