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

Explanation:

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

You are welcome:)

Answer 2

`(0.6)^(n)+n(0.4)(0.6)^(n-1)+(n(n-1)(0.4)^(2)0.6^(n-2))/(2) +(n(n-1)(n-2)0.4^(3)0.6^(n-3))/(6)`

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)`+
`(n(n-1)(p)^(2)q^(n-2))/(2)`+
`(n(n-1)(n-2)p^(3)q^(n-3))/(6)`

=
`(0.6)^(n)+n(0.4)(0.6)^(n-1)+[tex](n(n-1)(0.4)^(2)0.6^(n-2))/(2) +(n(n-1)(n-2)0.4^(3)0.6^(n-3))/(6)`