Respuesta :
Answer:
[tex](0.6)^{n}+n(0.4)(0.6)^{n-1}+\frac{n(n-1)(0.4)^{2}0.6^{n-2}}{2} +\frac{n(n-1)(n-2)0.4^{3}0.6^{n-3}}{6}[/tex]
Step-by-step explanation:
We will use the binomial distribution. Let X be the random variable representing the no. of boxes Hannah buys before betting a prize.
Our success is winning the prize, p =40/100 = 0.4
Then failure q = 1-0.4 = 0.6
Hannah keeps buying cereal boxes until she gets a prize. Then n be no. times she buys the boxes.
P(X ≤ 3) = P(X=0) +P(X=1)+P(X=2)+P(X=3)
= [tex]\binom{n}{0}p^{0}q^{n-0}[/tex] + [tex]\binom{n}{1}p^{1}q^{n-1}[/tex]+ [tex]\binom{n}{2}p^{2}q^{n-2}[/tex]+ [tex]\binom{n}{3}p^{3}q^{n-3}[/tex]
= [tex]q^{n}+npq^{n-1}[/tex]+[tex]\frac{n(n-1)(p)^{2}q^{n-2}}{2}[/tex]+[tex]\frac{n(n-1)(n-2)p^{3}q^{n-3}}{6}[/tex]
= [tex](0.6)^{n}+n(0.4)(0.6)^{n-1}+[tex]\frac{n(n-1)(0.4)^{2}0.6^{n-2}}{2} +\frac{n(n-1)(n-2)0.4^{3}0.6^{n-3}}{6}[/tex]
Answer:
Step-by-step explanation:
Use a random number generator ranging from 1-10......
You are welcome:)
