A coin is tossed n times. If the probability of getting at least two heads is greater than that of getting at least three tails by \(\frac{21}{128}\), then n is:

Calculation:

Probability of getting atleast two heads =  1 - (probability of getting zero head + probability of getting one head)

Probability of getting atleast three tails = 1 - (probability of getting zero tail + probability of getting one tail + probability of getting two tails)

Probability of getting zero head/tail = ⁿC₀(1/2)ⁿ

Probability of getting one head/tail = nC₁(1/2)n

Probability of getting two head/tails = nC₂(1/2)n

Probability of getting at least two heads =  1 - ( nC0(1/2)n +  nC1(1/2)n )

Probability of getting at least three tails =1 - ( nC0(1/2)n +  nC1(1/2)n +  nC2(1/2)n)

According to the question:

Probability of getting at least two heads - Probability of getting at least three tails = \(\frac{21}{128}\)

[1 - ( nC0(1/2)n +  nC1(1/2)n )] - [1 - ( nC0(1/2)n +  nC1(1/2)n +  nC2(1/2)n)] = \(\frac{21}{128}\)

⇒ nC2(1/2)n = \(\frac{21}{128}\)

⇒ \(\frac{n(n-1)}{2^n \times (1\times 2)} = \frac{21}{128}\).

for n = 7, the above equation satisfies.

The correct answer is option "1"