Question

The amount of money a gambler can win in a casino is determined by three independent rolls of a six-faced fair dice. The gambler wins Rs. 800 if he gets three sixes, Rs. 400 if he gets two sixes, and Rs. 100 in the event of getting only one six. The gambler does not win or lose any money in all other possible outcomes. The probability that a gambler will win at least Rs. 400 is ___________. (round off to 2 decimal places) 
 

Hint:

For dice or coin problems, always identify the random variable and use the Binomial distribution formula for independent identical trials.

Answer 1

Correct Answer: 0.41

Step 1: Define the random experiment.
There are 3 independent dice rolls. Let $X$ = number of sixes obtained. Possible values of $X$: 0, 1, 2, 3. Each die has a probability of $\dfrac{1}{6}$ for six and $\dfrac{5}{6}$ for not six.
Step 2: Identify favorable cases.
The gambler wins at least Rs. 400 if he gets two or three sixes. \[ P(X \ge 2) = P(X=2) + P(X=3) \]
Step 3: Use Binomial probability.
\[ P(X=k) = \binom{3}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{3-k} \] \[ P(X=2) = \binom{3}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right) = 3 \times \frac{1}{36} \times \frac{5}{6} = \frac{15}{216} \] \[ P(X=3) = \binom{3}{3}\left(\frac{1}{6}\right)^3 = \frac{1}{216} \]
Step 4: Add probabilities.
\[ P(X \ge 2) = \frac{15}{216} + \frac{1}{216} = \frac{16}{216} = 0.0741 \approx 0.07. \]
Step 5: Correction — check for Rs.400 or more. The problem asks for at least Rs.400, so indeed includes both $X=2$ and $X=3$.
Step 6: Conclusion.
\[ \boxed{P = 0.07} \] (round off to 2 decimal places)