Let X be the number of successes in five independent trials of a binomial experiment in which the probability of success is p = 3/5.
Find the following probabilities. (Round your answers to four decimal places.)
(a)
P(X = 4)
(b)
P(2 ≤ X ≤ 4)
a) .2592
To find the probabilities in a binomial experiment, we can utilize the binomial probability formula:
P(X = k) = (n C k) * p^k * q^(n-k)
where n is the number of trials, p is the probability of success, q is the probability of failure (1 - p), and (n C k) represents the combination formula which calculates the number of ways to choose k successes out of n trials.
In this case, n = 5, p = 3/5, and q = 1 - p = 2/5.
(a) To find P(X = 4):
P(X = 4) = (5 C 4) * (3/5)^4 * (2/5)^(5-4)
Using the combination formula (n C k) = n! / (k! * (n-k)!):
(5 C 4) = 5! / (4! * (5-4)!) = 5
Substituting the values into the formula:
P(X = 4) = 5 * (3/5)^4 * (2/5)^(5-4)
= 5 * (3/5)^4 * (2/5)^1
= 5 * (81/625) * (2/5)
= 5 * (162/3125)
= 810/3125
≈ 0.2592
Therefore, P(X = 4) is approximately 0.2592.
(b) To find P(2 ≤ X ≤ 4):
P(2 ≤ X ≤ 4) = P(X = 2) + P(X = 3) + P(X = 4)
To find P(X = 2):
P(X = 2) = (5 C 2) * (3/5)^2 * (2/5)^(5-2)
= (10) * (9/25) * (8/25)
= 720/625
= 1.152
To find P(X = 3):
P(X = 3) = (5 C 3) * (3/5)^3 * (2/5)^(5-3)
= (10) * (27/125) * (32/125)
= 864/625
= 1.3824
To find P(X = 4), we have already calculated it in part (a) as 0.2592.
Adding these probabilities:
P(2 ≤ X ≤ 4) = 1.152 + 1.3824 + 0.2592
= 2.7936
Therefore, P(2 ≤ X ≤ 4) is equal to 2.7936.
To find the probabilities in a binomial experiment, we need to use the binomial probability formula:
P(X = k) = (nCk) * p^k * (1 - p)^(n - k)
where:
- P(X = k) is the probability of having exactly k successes
- n is the number of trials
- k is the number of successes
- p is the probability of success in each trial
- (nCk) represents the number of combinations of n items taken k at a time, which can be calculated using the formula: n! / (k! * (n - k)!)
Now let's solve the problems:
(a) P(X = 4)
In this case, we have n = 5 (number of trials) and k = 4 (number of successes). The probability of success in each trial is p = 3/5.
Using the formula:
P(X = 4) = (5C4) * (3/5)^4 * (1 - 3/5)^(5 - 4)
To calculate (5C4), we have:
5! / (4! * (5 - 4)!) = 5
Plugging in the values:
P(X = 4) = 5 * (3/5)^4 * (2/5)^1 = 5 * (81/625) * (2/5) = 0.1296
Therefore, P(X = 4) is approximately 0.1296.
(b) P(2 ≤ X ≤ 4)
To find this probability, we need to sum the probabilities of having 2, 3, and 4 successes.
Using the formula:
P(2 ≤ X ≤ 4) = P(X = 2) + P(X = 3) + P(X = 4)
For P(X = 2):
P(X = 2) = (5C2) * (3/5)^2 * (1 - 3/5)^(5 - 2)
Calculating (5C2):
5! / (2! * (5 - 2)!) = 10
Plugging in the values:
P(X = 2) = 10 * (3/5)^2 * (2/5)^3 = 10 * (9/25) * (8/125) = 0.1728
For P(X = 3):
P(X = 3) = (5C3) * (3/5)^3 * (1 - 3/5)^(5 - 3)
Calculating (5C3):
5! / (3! * (5 - 3)!) = 10
Plugging in the values:
P(X = 3) = 10 * (3/5)^3 * (2/5)^2 = 10 * (27/125) * (4/25) = 0.1152
For P(X = 4), we already calculated it in part (a):
P(X = 4) = 0.1296
Adding the probabilities together:
P(2 ≤ X ≤ 4) = 0.1728 + 0.1152 + 0.1296 = 0.4176
Therefore, P(2 ≤ X ≤ 4) is approximately 0.4176.