Question

The probability of obtaining $4$ heads, $4$ tails when $8$ unbiased coins are tossed is :

• A. $\frac{1}{2}$
• B. $\frac{C_4^8}{2^8}$
• C. $\frac{C_0^8}{2^8}$
• D. $\frac{1}{4}$

Answer 1

Answer: (B)

$\frac{C_4^8}{2^8}$

When we toss $8$ unbiased coins; selection of exactly $4$ heads from $8$ coins $=C_4^8$; (remaining coins will obtain $4$ tails)

Total number of ways which we could get by arranging $8$ number of coins is $=2^8$
Probability of obtaining exactly $4$ heads and $\left(4\right)$ tails is $=\frac{C_4^8}{2^8}$