Assume the random variable X has a binomial distribution with the given probability of obtaining a suc the probability of obtaining

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Question:

Assume the random variable X has a binomial distribution with the given probability of obtaining a suc the probability of obtaining a success. Round your answer to four decimal places. P(X < 5), n = 8, p = 0.4 - Answer How to enter your answer (opens in new window)

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Answers

0.0579

Step-by-step explanation:

P(X≤2) = P(X=0) + P(X=1) + P(X=2)

P(X≤2) = ₅C₀ (0.8)⁰ (0.2)⁵ + ₅C₁ (0.8)¹ (0.2)⁴ + ₅C₂ (0.8)² (0.2)³

P(X≤2) = 0.00032 + 0.0064 + 0.0512

P(X≤2) = 0.0579

0.3446

Step-by-step explanation:

P(X≥2) = 1 − P(X<2)

P(X≥2) = 1 − P(X=0) − P(X=1)

P(X≥2) = 1 − ₆C₀ (0.2)⁰ (0.8)⁶ − ₆C₁ (0.2)¹ (0.8)⁵

P(X≥2) = 1 − 0.2621 − 0.3932

P(X≥2) = 0.3446

0.7893

Step-by-step explanation:


[tex]Assume the random variable x has a binomial distribution with the given probability of obtaining a s[/tex]

15/8

Step-by-step explanation:

[tex]P(X1)= 1-P(X \leq 1)= 1- [P(X=0) +P(X=1)][/tex]

And if we use the probability mass function we got:

[tex]P(X=0)=(4C0)(0.6)^0 (1-0.6)^{4-0}=0.0256[/tex]  

[tex]P(X=1)=(4C1)(0.6)^1 (1-0.6)^{4-1}=0.1536[/tex]  

And replacing we got:

[tex]P(X1) =1- [0.0256 +0.1536]= 0.8208[/tex]

Step-by-step explanation:

Let X the random variable of interest, on this case we now that:  

[tex]X \sim Binom(n=4, p=0.6)[/tex]  

The probability mass function for the Binomial distribution is given as:  

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]  

Where (nCx) means combinatory and it's given by this formula:  

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]  

We want to find the following probability:

[tex]P(X 1)[/tex]

And for this case we can use the complement rule and we got:

[tex]P(X1)= 1-P(X \leq 1)= 1- [P(X=0) +P(X=1)][/tex]

And if we use the probability mass function we got:

[tex]P(X=0)=(4C0)(0.6)^0 (1-0.6)^{4-0}=0.0256[/tex]  

[tex]P(X=1)=(4C1)(0.6)^1 (1-0.6)^{4-1}=0.1536[/tex]  

And replacing we got:

[tex]P(X1) =1- [0.0256 +0.1536]= 0.8208[/tex]

[tex]P(X=14)=(18C14)(0.8)^{14} (1-0.8)^{18-14}=0.2153[/tex]

[tex]P(X=15)=(18C15)(0.8)^{15} (1-0.8)^{18-15}=0.2297[/tex]

[tex]P(X=16)=(18C16)(0.8)^{16} (1-0.8)^{18-16}=0.1722[/tex]

[tex]P(X=17)=(18C17)(0.8)^{17} (1-0.8)^{18-17}=0.0811[/tex]

[tex]P(X=18)=(18C18)(0.8)^{18} (1-0.8)^{18-18}=0.0180[/tex]

And adding the values we got:

[tex]P(X \geq 14)= 0.2153 +0.2297+0.1722+0.0811+0.0180=0.7164[/tex]

Step-by-step explanation:

Assuming this question: P(X≥14), n=18, p=0.8

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Solution to the problem

Let X the random variable of interest, on this case we now that:  

[tex]X \sim Binom(n=18, p=0.8)[/tex]  

The probability mass function for the Binomial distribution is given as:  

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]  

Where (nCx) means combinatory and it's given by this formula:  

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]  

And we want this probability:

[tex]P(X \geq 14) = P(X=14) +P(X=15) +P(X=16)+P(X=17)+P(X=18)[/tex]

And we can find the individual proabilities using the probability mass function:

[tex]P(X=14)=(18C14)(0.8)^{14} (1-0.8)^{18-14}=0.2153[/tex]

[tex]P(X=15)=(18C15)(0.8)^{15} (1-0.8)^{18-15}=0.2297[/tex]

[tex]P(X=16)=(18C16)(0.8)^{16} (1-0.8)^{18-16}=0.1722[/tex]

[tex]P(X=17)=(18C17)(0.8)^{17} (1-0.8)^{18-17}=0.0811[/tex]

[tex]P(X=18)=(18C18)(0.8)^{18} (1-0.8)^{18-18}=0.0180[/tex]

And adding the values we got:

[tex]P(X \geq 14)= 0.2153 +0.2297+0.1722+0.0811+0.0180=0.7164[/tex]

P(x > 10) = 0.6981.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

In this question:

[tex]n = 14, p = 0.8[/tex]

P(x>10)

[tex]P(x 10) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14)[/tex]

In which

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 11) = C_{14,11}.(0.8)^{11}.(0.2)^{3} = 0.2501[/tex]

[tex]P(X = 12) = C_{14,12}.(0.8)^{12}.(0.2)^{2} = 0.2501[/tex]

[tex]P(X = 13) = C_{14,13}.(0.8)^{13}.(0.2)^{1} = 0.1539[/tex]

[tex]P(X = 14) = C_{14,14}.(0.8)^{14}.(0.2)^{0} = 0.0440[/tex]

[tex]P(x 10) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) = 0.2501 + 0.2501 + 0.1539 + 0.0440 = 0.6981[/tex]

So P(x > 10) = 0.6981.

P(X > 4) = 0.8059

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

In this question, we have that:

[tex]n = 8, p = 0.7[/tex]

We want:

[tex]P(X 4) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)[/tex]

In which

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 5) = C_{8,5}.(0.7)^{5}.(0.3)^{3} = 0.2541[/tex]

[tex]P(X = 6) = C_{8,6}.(0.7)^{6}.(0.3)^{2} = 0.2965[/tex]

[tex]P(X = 7) = C_{8,7}.(0.7)^{7}.(0.3)^{1} = 0.1977[/tex]

[tex]P(X = 8) = C_{8,8}.(0.7)^{8}.(0.3)^{0} = 0.0576[/tex]

Then

[tex]P(X 4) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.2541 + 0.2965 + 0.1977 + 0.0576 = 0.8059[/tex]

So

P(X > 4) = 0.8059

P(X < 3) = 0.7443

Step-by-step explanation:

We are given that the random variable X has a binomial distribution with the given probability of obtaining a success. Also, given n = 6, p = 0.3.

The above situation can be represented through Binomial distribution;

[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]

where, n = number of trials (samples) taken = 6

            r = number of success = less than 3

           p = probability of success which in our question is 0.3.

LET X = a random variable

So, it means X ~ [tex]Binom(n=6, p=0.3)[/tex]

Now, Probability that X is less than 3 = P(X < 3)

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

             = [tex]\binom{6}{0}0.3^{0} (1-0.3)^{6-0}+ \binom{6}{1}0.3^{1} (1-0.3)^{6-1}+ \binom{6}{2}0.3^{2} (1-0.3)^{6-2}[/tex]

             = [tex]1 \times 1 \times 0.7^{6} +6 \times 0.3^{1} \times 0.7^{5} +15 \times 0.3^{2} \times 0.7^{4}[/tex]

             = 0.11765 + 0.30253 + 0.32414 = 0.7443

Therefore, P(X < 3) = 0.7443.

P(X = 17) = 0.3002

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

In this problem we have that:

[tex]n = 18, p = 0.9[/tex]

We want P(X = 17). So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 17) = C_{18,17}.(0.9)^{17}.(0.1)^{1} = 0.3002[/tex]

P(X = 17) = 0.3002

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