The required probability of success is given by 0.98906500
Assume the random variable X has a binomial distribution with the given probability of obtaining a success.
P(X<5), n=6, p=0.3
In binomial distribution for number trials we are investigating the probability of getting a success remain the same.
n = number of trails,
p = probability of success
x = the number of success
p(x<5) = P(x=0)+ p(x=1) + p(x=2)+p(x=3)+p(x=4)
= [tex]\binom{6}{0}(0.3)^0(1-0.3)^6+\binom{6}{1}(0.3)^1(1-0.3)^5+\binom{6}{2}(0.3)^2(1-0.3)^4+\binom{6}{3}(0.3)^3(1-0.3)^3+\binom{6}{4}(0.3)^4(1-0.3)^2[/tex]
p(x<5) = 0.98906500
Thus the required probability of success is given by 0.98906500
Learn more about Binomial distribution here:
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