Can you guys explain the solution for this problem ?

**If the probability for each canon shoots the target is 60%, which of the following of numbers of canons can reach the probability at 99%?**

A) 3

B) 4

C) 5

D) 6

E) 7

Source?

The solution involves a logarithmic equation. Pretty sure such a question will NOT be asked. Anyway solved it till the equation.

For 1 cannon Prob= 0.6 and not hitting = 0.4

for 2 = 0.6^2

for 3 = 0.6^3

desired result: 1-0.4^n>=0.99

Kaplan Test 7

probability to not shoot the target = 1- 0.99 = 0.01

(1-0.6) ^ n <= 0.01

on choosing numbers, the above equation holds true at n > 5.

Isn’t it a fair question to ask by the ets too, the logic is the same. I have found a similar question in the GMAT club too