The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96, is?

The probability of getting exactly 0 heads in n tosses is = (1/2)^n
The probability of getting exactly 1 head in n tosses is nC1(1/2)^n
The probability of getting at least 2 heads = 1 - (1/2)^n - n(1/2)^n
The probability of getting at least 2 heads is 0.96.
Hence, 1 - (1/2)^n - n(1/2)^n < 0.04
1 - (1+n)(1/2)^n < 0.04
(1+n)(1/2)^n > 0.96
Let's try some values of n:
n = 1: (1+1)(1/2)^1 = 1 > 0.96
n = 2: (1+2)(1/2)^2 = 3/4 = 0.75 < 0.96
n = 3: (1+3)(1/2)^3 = 4/8 = 0.5 < 0.96
n = 4: (1+4)(1/2)^4 = 5/16 = 0.3125 < 0.96
n = 5: (1+5)(1/2)^5 = 6/32 = 0.1875 < 0.96
n = 6: (1+6)(1/2)^6 = 7/64 = 0.109375 < 0.96
n = 7: (1+7)(1/2)^7 = 8/128 = 0.0625 < 0.96
n = 8: (1+8)(1/2)^8 = 9/256 = 0.03515625 < 0.96
n = 9: (1+9)(1/2)^9 = 10/512 = 0.01953125 < 0.96
The least value of n for which (1+n)(1/2)^n > 0.96 is n=8. However, this inequality is actually reversed. It should be:
1 - (1/2)^n - n(1/2)^n ≥ 0.96
(1+n)(1/2)^n ≤ 0.04
Let's test values of n again:
n=8: (1+8)(1/2)^8 = 9/256 ≈ 0.035 > 0.04 (False)
n=9: (1+9)(1/2)^9 = 10/512 ≈ 0.0195 < 0.04 (True)
Therefore, the minimum number of tosses is 9.