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Sagot :
The binomial distribution is
[tex]\begin{gathered} P(X=k)=(nbinomialk)p^k(1-p)^{n-k} \\ k\rightarrow\text{ number of successful trials} \\ n\rightarrow\text{ total number of trials} \\ p\rightarrow\text{ probability of a trial being successful} \\ (nbinomialk)=\frac{n!}{(n-k)!k!} \end{gathered}[/tex]Therefore, in our case, the distribution is
[tex]n=20,p=20\%=0.2[/tex]And we are interested in the probability of k=0 (0 defective bulbs) and k=1 (1 defective bulb). Thus,
[tex]P(X=0)=(20binomial0)(0.2)^0(0.8)^{20}=1*1*0.8^{20}=0.8^{20}[/tex]Similarly,
[tex]P(X=1)=(20binomial1)(0.2)^1(0.8)^{19}=20*0.2*(0.8)^{19}=4*0.8^{19}[/tex]Hence,
[tex]P(X\leq1)=P(X=0)+P(X=1)=0.8^{20}+4(0.8)^{19}\approx0.0692[/tex]The probability of the 20 bulbs including 1 or fewer defective bulbs is 0.0692, the answer is 0.0692
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