A fair die is thrown twenty times. The probability that on the tenth throw the fourth six appears is:
Answer & explanation
Correct answer: option 3
Let success : getting '6' ; failure : no '6'
P(required) = ${^9C}_3(\frac{1}{6})^3(\frac{5}{6})^6.\frac{1}{6}=\frac{84×5^6}{6^{10}}$