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

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) ((3/4))10 (B) ((2/7))10 (C) ((1/4))10 (D) ((1/2))10

Tardigrade Admin · Accepted Answer

For any toss, total cases: HH,HT,TH,TT ( 4 cases) ∴ both of them will not get tail at same toss, then the favorable cases: HH,HT,TH ( 3 cases) Hence, the required probability =((3/4))10

Q. 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

$(\frac{3}{4})^{10}$

$(\frac{2}{7})^{10}$

$(\frac{1}{4})^{10}$

$(\frac{1}{2})^{10}$

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

Q. 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

(43​)10

(72​)10

(41​)10

(21​)10