Question:
A pair of dice is thrown independently \(3\) times. The probability of getting a total score of at least \(9\) twice, is:
A pair of dice is thrown independently \(3\) times. The probability of getting a total score of at least \(9\) twice, is:
Show 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.
• \(n\) = total trials
• \(r\) = number of successes
• \(p\) = probability of success
• \(q=1-p\)
This is the standard binomial distribution formula.
Updated On: Jun 17, 2026
- \( \dfrac{925}{5832} \)
- \( \dfrac{975}{5832} \)
- \( \dfrac{1025}{5832} \)
- \( \dfrac{1075}{5832} \)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
When an experiment is repeated independently several times and we count the number of successes, we use the binomial probability formula.
A success here means:
\[
\text{Sum on two dice} \geq 9
\]
Possible sums at least \(9\):
\[
9,10,11,12
\]
Step 1: Find probability of success in one throw.
Total outcomes when two dice are thrown: \[ 36 \] Now count favorable outcomes. For sum \(9\): \[ (3,6),(4,5),(5,4),(6,3) \] Total: \[ 4 \] For sum \(10\): \[ (4,6),(5,5),(6,4) \] Total: \[ 3 \] For sum \(11\): \[ (5,6),(6,5) \] Total: \[ 2 \] For sum \(12\): \[ (6,6) \] Total: \[ 1 \] Hence total favorable outcomes: \[ 4+3+2+1=10 \] Therefore, \[ p=\frac{10}{36}=\frac{5}{18} \] Probability of failure: \[ q=1-p = 1-\frac{5}{18} = \frac{13}{18} \]
Step 2: Use binomial probability formula.
We need probability of getting success exactly twice in \(3\) trials. Thus, \[ P(X=2) = {}^3C_2 p^2 q \] Substituting values: \[ = 3\left(\frac{5}{18}\right)^2\left(\frac{13}{18}\right) \] \[ = 3\times \frac{25}{324}\times \frac{13}{18} \] \[ = \frac{975}{5832} \] Thus, \[ \boxed{\frac{975}{5832}} \] Hence the correct option is: \[ \boxed{(B)} \]
Step 1: Find probability of success in one throw.
Total outcomes when two dice are thrown: \[ 36 \] Now count favorable outcomes. For sum \(9\): \[ (3,6),(4,5),(5,4),(6,3) \] Total: \[ 4 \] For sum \(10\): \[ (4,6),(5,5),(6,4) \] Total: \[ 3 \] For sum \(11\): \[ (5,6),(6,5) \] Total: \[ 2 \] For sum \(12\): \[ (6,6) \] Total: \[ 1 \] Hence total favorable outcomes: \[ 4+3+2+1=10 \] Therefore, \[ p=\frac{10}{36}=\frac{5}{18} \] Probability of failure: \[ q=1-p = 1-\frac{5}{18} = \frac{13}{18} \]
Step 2: Use binomial probability formula.
We need probability of getting success exactly twice in \(3\) trials. Thus, \[ P(X=2) = {}^3C_2 p^2 q \] Substituting values: \[ = 3\left(\frac{5}{18}\right)^2\left(\frac{13}{18}\right) \] \[ = 3\times \frac{25}{324}\times \frac{13}{18} \] \[ = \frac{975}{5832} \] Thus, \[ \boxed{\frac{975}{5832}} \] Hence the correct option is: \[ \boxed{(B)} \]
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