Question

A pair of dice is thrown independently \(3\) times. The probability of getting a total score of at least \(9\) twice, is:

• A. \( \dfrac{925}{5832} \)
• B. \( \dfrac{975}{5832} \)
• C. \( \dfrac{1025}{5832} \)
• D. \( \dfrac{1075}{5832} \)

Hint:

For repeated independent trials: \[ P(X=r)={}^nC_r p^r q^{n-r} \] where:

• \(n\) = total trials

• \(r\) = number of successes

• \(p\) = probability of success

• \(q=1-p\)
This is the standard binomial distribution formula.

Answer 1

The Correct Option is B

Step 1: Define a success.
A success is a total of at least $9$ on two dice, that is a sum of $9,10,11,$ or $12$.
Step 2: Count favourable sums.
Sum $9$: $4$ ways; sum $10$: $3$ ways; sum $11$: $2$ ways; sum $12$: $1$ way. Total $=10$ out of $36$.
Step 3: Find $p$ and $q$.
So $p=\dfrac{10}{36}=\dfrac{5}{18}$ and $q=1-p=\dfrac{13}{18}$.
Step 4: Set up the binomial.
Three throws, want exactly two successes: $P(X=2)={}^3C_2\,p^2q$.
Step 5: Substitute.
$=3\left(\dfrac{5}{18}\right)^2\left(\dfrac{13}{18}\right)=3\cdot\dfrac{25}{324}\cdot\dfrac{13}{18}$.
Step 6: Simplify.
This is $\dfrac{975}{5832}$. \[ \boxed{\frac{975}{5832}} \]