Two fair and ordinary dice are rolled simultaneously 4 times. The probability that both dice will show same outcome exactly twice, is equal to
Answer & explanation
Correct answer: option 1
Probability of showing equal outcomes in any specific roll of dice is $\frac{6}{36}$, i.e $\frac{1}{6}$
Thus required probability
= Getting exactly 2 successes in 4 trials.
$={ }^4 C_2 .\left(\frac{1}{6}\right)^2 .\left(\frac{5}{6}\right)^2=\frac{25}{216}$