Respuesta :
The given information is:
- p=0.05 (the probability of success)
- n=12 (the sample size)
We can apply the binomial distribution formula to find the probabilities. The formula is:
[tex]P(x)=\frac{n!}{(n-x)!x!}p^xq^{n-x}[/tex]q is equal to 1-p, so:
[tex]q=1-0.05=0.95[/tex]a. Find the probability that at least 3 people believe that they have seen a UFO:
This is equal to:
[tex]P(x\ge3)=1-(P(x=0)+P(x=1)+P(x=2))[/tex]Let's find the individual probabilities:
[tex]\begin{gathered} P(x=0)=\frac{12!}{(12-0)!1!}*0.05^00.95^{12-0} \\ P(x=0)=1*1*0.5404 \\ P(x=0)=0.5404 \\ \\ P(x=1)=\frac{12!}{(12-1)!1!}*0.05^10.95^{12-1} \\ P(x=1)=12*0.05*0.5688 \\ P(x=1)=0.341 \\ \\ P(x=2)=\frac{12!}{(12-2)!2!}*0.05^20.95^{12-2} \\ P(x=2)=66*0.0025*0.5987 \\ P(x=2)=0.099 \end{gathered}[/tex]So, the P(x>=3) is:
[tex]\begin{gathered} P(x\ge3)=1-(0.5404+0.341+0.099) \\ P(x\ge3)=0.0196 \end{gathered}[/tex]The probability that at least 3 people believe that they have seen a UFO is 0.0196.
b. P(3 or 4 people believe that they have seen a UFO):
This is given by:
[tex]P(3\leq x\leq4)=P(x=3)+P(x=4)[/tex]Let's find the individual probabilities:
[tex]\begin{gathered} P(x=3)=\frac{12!}{(12-3)!3!}*0.05^30.95^{12-3} \\ P(x=3)=220*0.0001*0.6303 \\ P(x=3)=0.0173 \\ \\ P(x=4)=\frac{12!}{(12-4)!4!}*0.05^40.95^{12-4} \\ P(x=4)=495*0.000006*0.663 \\ P(x=4)=0.0021 \\ \\ \text{ So:} \\ P(3\leq x\leq4)=0.0173+0.0021=0.0194 \end{gathered}[/tex]The probability that 3 or 4 people believe that they have seen a UFO is 0.0194.
c. P(exactly 2 people believe they have seen a UFO)
This is equal to:
[tex]\begin{gathered} P(x=2) \\ \text{ We already found it in part A, so:} \\ P(x=2)=0.099 \end{gathered}[/tex]The probability that exactly 2 people believe they have seen a UFO is 0.099.
