A fair coin is tossed 10 times. The probability of obtaining exactly six heads is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\frac{105}{512}$
P(exactly 6 heads)
$={^{10}C}_6×(\frac{1}{2})^6(\frac{1}{2})^4$
$=\frac{10!}{6!(10-6)1}=\frac{1}{2^{10}}$
$=\frac{10×9×8×7}{4×3×2}×\frac{1}{2^{10}}=\frac{105}{512}$