Question

A fair die is rolled 72 times and the percentage of 6s is recorded. What is the probability that at most 10% of the rolls are 6s

Answer 1

Answer:

P ($\hat p$ ≤ 0.10)

Step-by-step explanation:

The probability in terms of statistics for this given problem can be written as follows.

Let consider X to the random variable that represents the number of 6's in 7 throws of a dice, then:

X $\sim$Bin ( n = 72, p = 0.167)

E(X) = np

E(X) = 72× 0.167

E(X) = 12.024

E(X) $\simeq$  12

p+q =1

q = 1 - p

q = 1 - 0.167

q = 0.833

V(X) = npq

V(X) = 72 × 0.167 × 0.833

V(X) = 10.02

V(X) $\simeq$ 10

∴ X $\sim$ N ($\mu = 12, \sigma^2 =10$)

⇒ $\hat p = \dfrac{X}{n} \sim N ( p, \dfrac{pq}{n})$

where p = 0.167 and $\dfrac{pq}{n}$ = $\dfrac{0.167 \times 0.833}{72}$ = 0.00193

∴ P(at most 10% of rolls are 6's)

i.e

P ($\hat p$ ≤ 0.10)