Tamyra is making four cookies and has exactly four chocolate chips. If she

distributes the chips randomly into the four cookies, what is the probability that there are

no more than two chips in any one cookie? Express your answer as a common fraction.

The probability that there are no more than two chips in any one cookie = 23/25

Step-by-step explanation:

The mean number of chocolate chips per cookie is 4/4 = 1 chocolate chip per cookie.

Using the Poisson distribution formula

P(X = x) = (e^-λ)(λˣ)/x!

Mean = λ = 1

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

For P(X = 0),

P(X = 0) = (e⁻¹)(1⁰)/(0!) = 0.36788

For P(X = 1)

P(X = 1) = (e⁻¹)(1¹)/(1!) = 0.36788

For P(X = 2)

P(X = 2) = (e⁻¹)(1²)/(2!) = 0.18394

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

P(X ≤ 2) = 0.36788 + 0.36788 + 0.18394 = 0.9197 ≈ 0.92 = 23/25

cross multiplying, we get :

⇒ (3) × (5) = (9) × (x)

⇒ 15 = 9x

the formula for converting absorbance to transmittance goes as:

a = log (1/t)  a = 2 - log(%t)  10^a = 1/t  t = 10^-a  %t = 100(10^-a)

so at the given value of a,   a = 1.5

%t = 100(10^-1.5)

%t = 3.16%