For a binomial distribution with parameters
n = 5 , p = 0.3 . Find the probabilities of
getting :
(i) Atleast 3 successes.
(ii) Atmost 3 successes.
n = 5 , p = 0.3
binomial coefficients for n = 5:
1 5 10 10 5 1
p = .3
1-p = .7
P(0) = .3^0 (.7)^5 = .168
p(1) = 5 (.3)^1(.7)^4 = .360
P(2) = 10 (.3)^2(.7)^3 = .309
P(3) = 10 (.3)^3(.7)^2 = .1323
P(4) = 5 (.3)^4(.7)^1 = .028
P(5) = 1 (.3)^5(.7)^0 = .002
first answer is P(3)+P(4)+P(5)
second answer is P(0)+P(1)+P(2)+P(3)
Why did the binomial distribution go to therapy? Because it had too many parameters!
(i) To find the probability of getting at least 3 successes, we need to calculate the probability of getting 3, 4, and 5 successes and add them up. So, let's do some math:
P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X = 5)
P(X = 3) = (5 choose 3) * (0.3)^3 * (0.7)^2
P(X = 4) = (5 choose 4) * (0.3)^4 * (0.7)^1
P(X = 5) = (5 choose 5) * (0.3)^5 * (0.7)^0
Now, let's calculate it:
P(X = 3) = 10 * 0.027 * 0.49
P(X = 4) = 5 * 0.0081 * 0.7
P(X = 5) = 1 * 0.000243 * 1
Now, sum them up:
P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X = 5)
= 0.1377 + 0.0357 + 0.000243
So, the probability of getting at least 3 successes is approximately 0.1736, or about 17.36%.
(ii) To find the probability of getting at most 3 successes, we need to calculate the probability of getting 0, 1, 2, and 3 successes and add them up. Let's clown around with the math again:
P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
P(X = 0) = (5 choose 0) * (0.3)^0 * (0.7)^5
P(X = 1) = (5 choose 1) * (0.3)^1 * (0.7)^4
P(X = 2) = (5 choose 2) * (0.3)^2 * (0.7)^3
P(X = 3) = (5 choose 3) * (0.3)^3 * (0.7)^2
Calculating:
P(X = 0) = 1 * 1 * 0.16807
P(X = 1) = 5 * 0.3 * 0.2401
P(X = 2) = 10 * 0.09 * 0.343
P(X = 3) = 10 * 0.027 * 0.49
Now, let's sum them up:
P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 0.16807 + 0.36015 + 0.3087 + 0.1377
So, the probability of getting at most 3 successes is approximately 0.9746, or about 97.46%.
Remember, laughter is the best way to deal with distributions!
To find the probabilities of getting at least 3 successes and at most 3 successes, we can use the binomial probability formula:
P(x) = C(n, x) * p^x * (1 - p)^(n - x)
Where:
P(x) = Probability of getting x successes
C(n, x) = Combination of n choose x (denotes the number of ways to select x items from a set of n items)
(i) At least 3 successes:
To find the probability of getting at least 3 successes, we need to calculate the probability of getting 3, 4, and 5 successes and add them together.
P(at least 3 successes) = P(3 successes) + P(4 successes) + P(5 successes)
P(x) = C(n, x) * p^x * (1 - p)^(n - x)
P(3 successes) = C(5, 3) * 0.3^3 * (1 - 0.3)^(5 - 3)
= 10 * 0.3^3 * 0.7^2
= 0.3087
P(4 successes) = C(5, 4) * 0.3^4 * (1 - 0.3)^(5 - 4)
= 5 * 0.3^4 * 0.7^1
= 0.1323
P(5 successes) = C(5, 5) * 0.3^5 * (1 - 0.3)^(5 - 5)
= 1 * 0.3^5 * 0.7^0
= 0.00243
P(at least 3 successes) = 0.3087 + 0.1323 + 0.00243
= 0.4434
Therefore, the probability of getting at least 3 successes is 0.4434.
(ii) At most 3 successes:
To find the probability of getting at most 3 successes, we need to calculate the probability of getting 0, 1, 2, and 3 successes and add them together.
P(at most 3 successes) = P(0 successes) + P(1 success) + P(2 successes) + P(3 successes)
P(0 successes) = C(5, 0) * 0.3^0 * (1 - 0.3)^(5 - 0)
= 1 * 1 * 0.7^5
= 0.1681
P(1 success) = C(5, 1) * 0.3^1 * (1 - 0.3)^(5 - 1)
= 5 * 0.3^1 * 0.7^4
= 0.36015
P(2 successes) = C(5, 2) * 0.3^2 * (1 - 0.3)^(5 - 2)
= 10 * 0.3^2 * 0.7^3
= 0.3087
P(3 successes) = C(5, 3) * 0.3^3 * (1 - 0.3)^(5 - 3)
= 10 * 0.3^3 * 0.7^2
= 0.1323
P(at most 3 successes) = 0.1681 + 0.36015 + 0.3087 + 0.1323
= 0.96925
Therefore, the probability of getting at most 3 successes is 0.96925.
To find the probabilities of getting at least 3 successes and at most 3 successes for a binomial distribution with parameters n = 5 and p = 0.3, we can use the formula for the binomial probability mass function:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Where:
P(X = k) is the probability of getting exactly k successes.
C(n, k) is the number of combinations of n items taken k at a time, defined as C(n, k) = n! / (k! * (n - k)!).
p is the probability of success in a single trial.
(1 - p) is the probability of failure in a single trial.
Now let's calculate the probabilities:
(i) At least 3 successes:
In this case, we need to find the probabilities of getting 3, 4, and 5 successes and sum them up.
P(X >= 3) = P(X = 3) + P(X = 4) + P(X = 5)
P(X = 3) = C(5, 3) * (0.3)^3 * (0.7)^2
P(X = 4) = C(5, 4) * (0.3)^4 * (0.7)^1
P(X = 5) = C(5, 5) * (0.3)^5 * (0.7)^0
Calculate each term using the combination formula and simplify the expression.
C(5, 3) = 5! / (3! * (5 - 3)!) = 10
C(5, 4) = 5! / (4! * (5 - 4)!) = 5
C(5, 5) = 5! / (5! * (5 - 5)!) = 1
Now plug into the formula:
P(X >= 3) = 10 * (0.3)^3 * (0.7)^2 + 5 * (0.3)^4 * (0.7)^1 + 1 * (0.3)^5 * (0.7)^0
P(X >= 3) = 0.1323 + 0.02835 + 0.00243
P(X >= 3) = 0.16308
Therefore, the probability of getting at least 3 successes is approximately 0.16308.
(ii) At most 3 successes:
In this case, we need to find the probabilities of getting 0, 1, 2, and 3 successes and sum them up.
P(X <= 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
P(X = 0) = C(5, 0) * (0.3)^0 * (0.7)^5
P(X = 1) = C(5, 1) * (0.3)^1 * (0.7)^4
P(X = 2) = C(5, 2) * (0.3)^2 * (0.7)^3
P(X = 3) = C(5, 3) * (0.3)^3 * (0.7)^2
Calculate each term using the combination formula and simplify the expression.
C(5, 0) = 5! / (0! * (5 - 0)!) = 1
C(5, 1) = 5! / (1! * (5 - 1)!) = 5
C(5, 2) = 5! / (2! * (5 - 2)!) = 10
C(5, 3) = 5! / (3! * (5 - 3)!) = 10
Now plug into the formula:
P(X <= 3) = 1 * (0.3)^0 * (0.7)^5 + 5 * (0.3)^1 * (0.7)^4 + 10 * (0.3)^2 * (0.7)^3 + 10 * (0.3)^3 * (0.7)^2
P(X <= 3) = 0.16807 + 0.36015 + 0.3087 + 0.1323
P(X <= 3) = 0.96922
Therefore, the probability of getting at most 3 successes is approximately 0.96922.