In each part, assume the random variable X has a binomial distribution with the given parameters. Compute the probability of the event.
(a) n=3,p=0.7
Pr(X=3)=
(b) n=6,p=0.6
Pr(X=6)=
(c) n=6,p=0.2
Pr(X=5)=
(d) n=5,p=0.8
Pr(X=0)=
(c) n=6,p=0.2
Pr(X=5)=
interpretation:
P(X=5) = C(6,5) (.2)^5 (.8)^1
= 6(.00032)(.8)
= .001536
(a) To compute Pr(X=3), we use the formula for the probability mass function of a binomial distribution given by Pr(X=k) = (n choose k) * p^k * (1-p)^(n-k).
Plugging in the values n=3 and p=0.7, we have Pr(X=3) = (3 choose 3) * 0.7^3 * (1-0.7)^(3-3) = 1 * 0.343 * 1 = 0.343.
So, the probability of getting a result of 3 in three trials with a success probability of 0.7 is 0.343.
(b) Similarly, for Pr(X=6) with n=6 and p=0.6, we have Pr(X=6) = (6 choose 6) * 0.6^6 * (1-0.6)^(6-6) = 1 * 0.046656 * 1 = 0.046656.
Therefore, the probability of obtaining a result of 6 in six trials with a success probability of 0.6 is approximately 0.046656.
(c) For Pr(X=5) with n=6 and p=0.2, we have Pr(X=5) = (6 choose 5) * 0.2^5 * (1-0.2)^(6-5) = 6 * 0.000064 * 0.8 = 0.0003072.
Hence, the probability of getting a result of 5 in six trials with a success probability of 0.2 is approximately 0.0003072.
(d) Lastly, for Pr(X=0) with n=5 and p=0.8, we have Pr(X=0) = (5 choose 0) * 0.8^0 * (1-0.8)^(5-0) = 1 * 1 * 0.00032 = 0.00032.
In conclusion, the probability of obtaining a result of 0 in five trials with a success probability of 0.8 is 0.00032.
To compute the probability in each part, we will use the formula for the probability mass function of a binomial distribution:
P(X=k) = C(n,k) * p^k * (1-p)^(n-k)
where:
- P(X=k) represents the probability that the random variable X takes the value k,
- n represents the number of trials,
- p represents the probability of success in each trial,
- C(n,k) represents the number of combinations of n items taken k at a time,
- ^ denotes exponentiation.
Let's calculate the probabilities:
(a) n = 3, p = 0.7
P(X=3) = C(3,3) * 0.7^3 * (1-0.7)^(3-3)
= 1 * 0.7^3 * 0.3^0
= 1 * 0.7^3 * 1
= 0.7^3
≈ 0.343
(b) n = 6, p = 0.6
P(X=6) = C(6,6) * 0.6^6 * (1-0.6)^(6-6)
= 1 * 0.6^6 * 0.4^0
= 1 * 0.6^6 * 1
= 0.6^6
≈ 0.046
(c) n = 6, p = 0.2
P(X=5) = C(6,5) * 0.2^5 * (1-0.2)^(6-5)
= 6 * 0.2^5 * 0.8^1
= 6 * 0.2^5 * 0.8
≈ 0.049
(d) n = 5, p = 0.8
P(X=0) = C(5,0) * 0.8^0 * (1-0.8)^(5-0)
= 1 * 0.8^0 * 0.2^5
= 1 * 1 * 0.2^5
= 0.2^5
= 0.00032
Therefore, the probabilities are:
(a) Pr(X=3) ≈ 0.343
(b) Pr(X=6) ≈ 0.046
(c) Pr(X=5) ≈ 0.049
(d) Pr(X=0) = 0.00032
To compute the probability of an event for a binomial distribution, we can use the formula:
Pr(X = k) = (n choose k) * (p^k) * ((1-p)^(n-k))
Where n is the number of trials, p is the probability of success, k is the number of successes, and (n choose k) represents the number of ways to choose k successes out of n trials.
Now, let's calculate the probabilities for each part:
(a) n=3, p=0.7, Pr(X=3)
Using the formula, we have:
Pr(X=3) = (3 choose 3) * (0.7^3) * ((1-0.7)^(3-3))
= 1 * 0.7^3 * (0.3^0)
= 0.7^3
= 0.343
Therefore, the probability Pr(X=3) is 0.343.
(b) n=6, p=0.6, Pr(X=6)
Pr(X=6) = (6 choose 6) * (0.6^6) * ((1-0.6)^(6-6))
= 1 * 0.6^6 * (0.4^0)
= 0.6^6
= 0.046656
Therefore, the probability Pr(X=6) is approximately 0.046656.
(c) n=6, p=0.2, Pr(X=5)
Pr(X=5) = (6 choose 5) * (0.2^5) * ((1-0.2)^(6-5))
= 6 * 0.2^5 * (0.8^1)
= 6 * 0.2^5 * 0.8
= 0.078336
Therefore, the probability Pr(X=5) is approximately 0.078336.
(d) n=5, p=0.8, Pr(X=0)
Pr(X=0) = (5 choose 0) * (0.8^0) * ((1-0.8)^(5-0))
= 1 * 0.8^0 * (0.2^5)
= 0.2^5
= 0.00032
Therefore, the probability Pr(X=0) is 0.00032.