$6$ ordinary dice are rolled. The probability that at least half of them will show at least $3$ is
- $\displaystyle 41 \times \frac {2^4}{3^6}$
- $\displaystyle \frac {2^4}{3^6}$
- $\displaystyle 20 \times \frac {2^4}{3^6}$
-
none of these
Probability of showing at least 3 is 4/6 = 2/3. Let p = 2/3, q = 1/3. We need at least 3 successes in 6 trials (Binomial distribution). P(X >= 3) = sum from k=3 to 6 of 6Ck * (2/3)^k * (1/3)^(6-k).
The probability of a single die showing at least 3 is 4/6, which equals 2/3. For at least half of the six dice to show at least 3, we must find the probability of exactly 3, exactly 4, exactly 5, or exactly 6 successes using the binomial distribution. We calculate this as 6C3(2/3)^3(1/3)^3 + 6C4(2/3)^4(1/3)^2 + 6C5(2/3)^5(1/3)^1 + 6C6(2/3)^6(1/3)^0. Combining these terms over a common denominator of 3^6 yields (160 + 240 + 192 + 64) / 729, which equals 656/729. Factoring out 2^4 from the numerator gives 41 x 2^4 / 3^6.