In a binomial distribution B(n, p = 1/4), if the probability of at least one success is greater than or equal to 9/10, then n is greater than

Given p = 1/4, q = 1 - p = 3/4
P(x ≥ 1) ≥ 9/10
1 - P(x = 0) ≥ 9/10
1 - ⁿC₀ * p⁰ * qⁿ ≥ 9/10
1 - qⁿ ≥ 9/10
1 - (3/4)ⁿ ≥ 9/10
(3/4)ⁿ ≤ 1 - 9/10
(3/4)ⁿ ≤ 1/10
Taking logarithm on both sides:
log((3/4)ⁿ) ≤ log(1/10)
n log(3/4) ≤ log(1/10)
n (log 3 - log 4) ≤ (log 1 - log 10)
n (log 3 - log 4) ≤ -1
n (log 3 - 2log 2) ≤ -1
n ≤ -1 / (log 3 - 2log 2)
n ≥ 1 / (2log 2 - log 3)
Approximating the values:
log 2 ≈ 0.3010
log 3 ≈ 0.4771
Therefore,
n ≥ 1 / (2(0.3010) - 0.4771)
n ≥ 1 / (0.6020 - 0.4771)
n ≥ 1 / 0.1249
n ≥ 8.0064
Hence n ≥ 9