probability on chips?

Question

A box contains 20 chips. There are three chips marked with a number 3, nine chips marked with a number 9, and eight chips marked with a number 27. If 100 chips are selected randomly with replacement, find the approximate probability that the product of the numebrs observed will be between 3^210 and 3^230.

Answer 1

Look at the exponents, you have 3¹,3², 3³.

you have three chips for exponent 1
you have nine chips for exponent 2
you have eight chips for exponent 3

you need to find the probability that the sum of 100 draws is between 210 and 230.

Find P[210 ≤ ΣX ≤ 230]

the central limit theorem stats the the sum, or mean, of any random variable will become approximately normal for large sample sizes.

ΣX ~ Normal(μ = 225, σ² =48.75)

μ = 100* [P(X=1) * 1 + P(X =2) * 2 + P(X = 3) * 3] = 225,
σ² = 100 * [P( X=1 ) * ( 1-2.25 )² + P( X=2 ) * ( 2-2.25 )² + + P( X=3 ) * ( 3-2.25 )²] = 48.75

P[210 ≤ ΣX ≤ 230]
= P[(210 - 225) / Sqrt(48.75) ≤ Z ≤ (230 - 225) / Sqrt(48.75) ]
= P(-2.15 < Z < 0.72]
= P( Z < 0.72) - P(Z < -2.15)
= 0.7642 - 0.0158
= 0.7484

where Z is the standard normal and the values come from the standard normal cdf table.

Answer 2

Relabel the chips with the numbers 1, 2 and 3.

Note that if chips are drawn with replacement and the (original) numbers multiplied by each other, the resultant product is as follows:

3^i * 3^j where i and j are elements of the set {3,9,27}

This product is equivalent to 3^(i+j).

However, note that 3 = 3^1, 9 = 3^2, 27 = 3^3

So the above question is equivalent to the following:

A box contains 20 chips. There are three chips marked with a number 1, nine chips marked with a number 2, and eight chips marked with a number 3. If 100 chips are selected randomly with replacement, find the approximate probability that the *SUM* of the numebrs observed will be between 210 and 230.

This is a relatively simple problem now to solve. Good luck.