Question

The probability that in 10 throws of a fair die, a score which is a multiple of 3 is obtained in 8 of the throws is:

• A. \( \frac{160}{3^{10}} \)
• B. \( \frac{140}{3^{10}} \)
• C. \( \frac{180}{3^{10}} \)
• D. \( \frac{100}{3^{10}} \)

Hint:

For “exactly r successes in n trials” problems, directly use binomial distribution: - Identify success and failure. - Find \(p\) and \(q\). - Use \( {}^nC_r p^r q^{n-r} \). - Simplify carefully after substitution.

Answer 1

The Correct Option is C

Concept: This is a Binomial Probability problem. The binomial probability formula is: \[ P(X=r)={}^{n}C_r p^r q^{n-r} \] where,

• \(n\) = total number of trials
• \(r\) = number of successes
• \(p\) = probability of success
• \(q=1-p\)

Here, Success = getting a multiple of 3. Multiples of 3 on a die are: \[ 3,6 \] So, \[ p=\frac{2}{6}=\frac13 \] and \[ q=1-\frac13=\frac23 \]
Step 1: Identify values for binomial distribution. Number of throws: \[ n=10 \] Number of successes: \[ r=8 \] Probability of success: \[ p=\frac13 \] Probability of failure: \[ q=\frac23 \]
Step 2: Apply binomial probability formula. \[ P(X=8)= {}^{10}C_8 \left(\frac13\right)^8 \left(\frac23\right)^2 \]
Step 3: Evaluate the combination and simplify. \[ {}^{10}C_8=\frac{10!}{8!2!}=45 \] Substituting: \[ P=45\left(\frac13\right)^8\left(\frac23\right)^2 \] \[ =45\cdot \frac{1}{3^8}\cdot \frac{4}{9} \] \[ =45\cdot \frac{4}{3^{10}} \] \[ =\frac{180}{3^{10}} \]