Respuesta :
Answer:
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.
Using the binomial distribution, it is found that P(X < 3) = 0.7442.
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Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, with p probability.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
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- In this question, we have that p = 0.3, n = 6.
- The desired probability is:
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{6,0}.(0.3)^{0}.(0.7)^{6} = 0.1176[/tex]
[tex]P(X = 1) = C_{6,1}.(0.3)^{1}.(0.7)^{5} = 0.3025[/tex]
[tex]P(X = 2) = C_{6,2}.(0.3)^{2}.(0.7)^{4} = 0.3241[/tex]
Then
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1176 + 0.3025 + 0.3241 = 0.7442[/tex]
A similar problem is given at https://brainly.com/question/15557838
