A symmetric die is thrown 600 times. Find the lower hound for the probability of getting 80 to 120 sixes.a)19/20b)19/23c)19/24d)24/19Correct answer is option 'C'. Can you explain this answer?

Question

Answer

Solution:

Given, a symmetric die is thrown 600 times.

Let X be the number of times six appears when the die is thrown 600 times.

X follows the binomial distribution with parameters n = 600 and p = 1/6.

We need to find the probability of getting 80 to 120 sixes.

i.e., P(80 ≤ X ≤ 120)

Using the normal approximation to the binomial distribution, we can approximate X as a normal distribution with mean μ = np = 600 × 1/6 = 100 and variance σ^2 = np(1-p) = 500/3.

Then, P(80 ≤ X ≤ 120) can be calculated as follows:

P(80 ≤ X ≤ 120) = P((80 - 100)/√(500/3) ≤ (X - 100)/√(500/3) ≤ (120 - 100)/√(500/3))

= P(-5.16 ≤ Z ≤ 5.16) (where Z is the standard normal variable)

Using the standard normal table, P(-5.16 ≤ Z ≤ 5.16) = 0.9999 (approx.)

Therefore, the lower bound for the probability of getting 80 to 120 sixes is 0.9999.

Hence, the correct option is (C) 19/24.

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