Question

40% of Oatypop cereal boxes contain a prize. Hannah plans to keep buying cereal until she gets a prize. What is the probability that Hannah only to buy 3 or less boxes before getti g a prize? We need to design a simulation. Which random can we use to best represent this situation

Answer 1

Answer:

$(0.6)^{n}+n(0.4)(0.6)^{n-1}+\frac{n(n-1)(0.4)^{2}0.6^{n-2}}{2} +\frac{n(n-1)(n-2)0.4^{3}0.6^{n-3}}{6}$

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)

             = $\binom{n}{0}p^{0}q^{n-0}$ +  $\binom{n}{1}p^{1}q^{n-1}$+ $\binom{n}{2}p^{2}q^{n-2}$+ $\binom{n}{3}p^{3}q^{n-3}$

             =  $q^{n}+npq^{n-1}$+$\frac{n(n-1)(p)^{2}q^{n-2}}{2}$+$\frac{n(n-1)(n-2)p^{3}q^{n-3}}{6}$

             = $(0.6)^{n}+n(0.4)(0.6)^{n-1}+[tex]\frac{n(n-1)(0.4)^{2}0.6^{n-2}}{2} +\frac{n(n-1)(n-2)0.4^{3}0.6^{n-3}}{6}$

Answer 2

Answer:

Step-by-step explanation:

Use a random number generator ranging from 1-10......

You are welcome:)