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Sagot :
Answer:0.849185
Step-by-step explanation:
Binomial probability formula we will use:
[tex]$P(x)=\frac{n !}{x !(n-x) !} p^{x} q^{n-x}$[/tex]
where,
[tex]n=5$\\$\mathrm{P}$ (probability of success) $=0.3$\\q=1-p[/tex]
The computation of the probability will be :
[tex]$P(x \leq 2 ; 5,0.3)=P(x=0 ; 5,0.3)+P(x=1 ; 5,0.3)+P(x=2 ; 5,0.3)+P(x=3 ; 5,0.3)$\\$=\left[\frac{5 !}{5 !(5-0) !}(0.3)^{0}(1-0.3)^{5-0}\right]+\left[\frac{5 !}{5 !(5-1) !}(0.3)^{1}(1-0.3)^{5-1}\right]$+\left[\frac{5 !}{5 !(5-2) !}(0.3)^{2}(1-0.3)^{5-2}\right]$++\left[\frac{5 !}{5 !(5-3) !}(0.3)^{3}(1-0.3)^{5-3}\right]$++\left[\frac{5 !}{5 !(5-4) !}(0.3)^{4}(1-0.3)^{5-4}\right]$\\$=0.1681+0.3601+0.3087+0.006615+0.00567$\\$\Rightarrow 0.849185$[/tex]
The probability that at most 4 are accepted = 0.849185
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