What is the probability of getting five heads and seven tails in 12 flips of a balanced coin?

Options

A. `C(12, 5)//(2^(5))`
B. `C(12, 5)//(2^(7))`
C. `C(12, 5)//(2^(12))`
D. `C(12, 7)//(2^(6))`

Answer

C

Explanation

Number of ways of selecting 5 heads cut of total 12 flips `= ""^(12)C_(5)`.
Probability of getting one head in a coin `= 1/2`
Also, probability of getting one tail in a coin `= 1/2`
Probability of getting 5 head `= (1/2)^(5)`
Probability of getting 7 tails `= (1/2)^(7)`
So, required probability
`= ""^(12)C_(5) (1/2)^(5) (1/2)^(7) = ""^(12)C_(5) (1/2)^(12) = (""^(12)C_(5))/(2^(12))`