Question

Each of $A$ and $B$ tosses a fair coin simultaneously $10$ times. The probability that both of them will not get tail at the same toss is equal to

• A. $\left(\frac{3}{4}\right)^{10}$
• B. $\left(\frac{2}{7}\right)^{10}$
• C. $\left(\frac{1}{4}\right)^{10}$
• D. $\left(\frac{1}{2}\right)^{10}$

Answer 1

Answer: (A)

For any toss, total cases: $HH,HT,TH,TT$ ( $4$ cases)
$\therefore $ both of them will not get tail at same toss, then the favorable cases: $HH,HT,TH$ ( $3$ cases)
Hence, the required probability $=\left(\frac{3}{4}\right)^{10}$