An experiment succeeds twice as often as it fails. The probability that in the next six trials there will be at least four successes, is:
1. \(\rm \dfrac{240}{729}\)
2. \(\rm \dfrac{496}{729}\)
3. \(\rm \dfrac{220}{729}\)
4. \(\rm \dfrac{233}{729}\)
1. \(\rm \dfrac{240}{729}\)
2. \(\rm \dfrac{496}{729}\)
3. \(\rm \dfrac{220}{729}\)
4. \(\rm \dfrac{233}{729}\)
Correct Answer – Option 2 : \(\rm \dfrac{496}{729}\)
Concept:
Binomial Distribution:
P(X = k) = nCk pk (1 – p)n – k
Calculation:
For every 3 trials, the experiment succeeds 2 times and fails 1 time.
∴ Probability of success: p = \(\dfrac23\).
For at least 4 success in 6 trials, n = 6 and k = 4, 5, 6. The required probability is therefore:
\(\rm ^6C_4\left(\dfrac23\right)^4\left(1-\dfrac23\right)^{6-4}+^6C_5\left(\dfrac23\right)^5\left(1-\dfrac23\right)^{6-5}+^6C_6\left(\dfrac23\right)^6\left(1-\dfrac23\right)^{6-6}\)
= \(\rm (15)\left(\dfrac23\right)^4\left(\dfrac13\right)^2+(6)\left(\dfrac23\right)^5\left(\dfrac13\right)^1+(1)\left(\dfrac23\right)^6\left(\dfrac13\right)^0\)
= \(\rm \dfrac{240}{729}+\dfrac{192}{729}+\dfrac{64}{729}\)
= \(\rm \dfrac{496}{729}\).