If the probability that event A will happen is 1/4 and the probability of A not happening is 3/4, find the probability that event A happens 3 times out of 7 experiments
A binomial distribution applies when
1. Trials have exactly two possible outcomes
2. probabilities remain constant throughout trials.
3. there is a defined number of trials.
4. trials are independent of each other.
5. the random variable is the number of successes.
Since the problem appears to satisfy all the requirements, we will use a binomial distribution to model the probabilities.
p=P(A)=1/4
q=1-P(A)=3/4
From (p+q)^7, we look for the term
P(3)
=(7,3)p^3*q^4
=(7!/(3!4!))(1/4)^3(3/4)^4
=2835/16384
To find the probability that event A happens 3 times out of 7 experiments, we can use the concept of binomial probability.
The probability of event A happening in a single experiment is given as 1/4. Therefore, the probability of A not happening in a single experiment is 1 - 1/4 = 3/4.
Now, let's calculate the probability of A happening exactly 3 times out of 7 experiments.
The formula for the binomial probability is:
P(X=k) = (n C k) * p^k * q^(n-k)
Where:
- P(X=k) is the probability of getting k successes in n trials.
- (n C k) is the binomial coefficient, which represents the number of ways to choose k items from a set of n items. It is calculated by the formula: (n C k) = n! / (k! * (n-k)!), where "!" denotes the factorial.
- p is the probability of success in a single trial.
- q is the probability of failure in a single trial, which is equal to 1 - p.
- k is the number of successes we want.
- n is the total number of trials or experiments.
In this case, we want to find the probability of A happening exactly 3 times out of 7 experiments. Therefore, k = 3, n = 7, p = 1/4, and q = 3/4.
Using the formula, we can calculate the probability as follows:
P(X=3) = (7 C 3) * (1/4)^3 * (3/4)^(7-3)
Let's calculate each part separately:
- (7 C 3) = 7! / (3! * (7-3)!) = (7 * 6 * 5) / (3 * 2 * 1) = 35
- (1/4)^3 = 1/4 * 1/4 * 1/4 = 1/64
- (3/4)^(7-3) = (3/4)^4 = 81/256
Now, let's substitute these values back into the formula:
P(X=3) = 35 * (1/64) * (81/256)
P(X=3) = (35 * 1 * 81) / (64 * 256)
P(X=3) = 2835 / 16384
Therefore, the probability that event A happens 3 times out of 7 experiments is 2835/16384, which is approximately 0.173.