Four dice are thrown simultaneously. Find the probability that two of them show the same face and remaining two show the different faces.
Solution
Correct Answer: Option B
Select a number which ocurs on two dice out of six numbers (1, 2, 3, 4, 5, 6). This can be done in \({}^6C_1\) ways.
Now select two distinct number out of remaining 5 numbers which can be done in \({}^5C_2\) ways. Thus these 4 numbers can be arranged in 4!/2! ways.
So, the number of ways in which two dice show the same face and the remaining two show different faces is
\({}^6C_1\times{}^5C_2\times\frac{4!}{2!}=720\)
=> n(E) = 720
= \(\therefore P\left(E\right)=\frac{720}{6^4}=\frac59\)
Practice More Questions on Our App!
Download our app for free and access thousands of MCQ questions with detailed solutions