1.

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. \(\rm \dfrac{233}{729}\)

Answer» Correct Answer - Option 2 : \(\rm \dfrac{496}{729}\)

Concept:

Binomial Distribution:

  • A Binomial Distribution is an experiment in which only ONE OUT OF TWO outcomes is possible. For example: Head or Tail, Yes or No, 1 or 0, etc.
  • If ‘n’ and ‘p’ are the parameters, then ‘n’ denotes the total number of times the experiment is conducted and ‘p’ denotes the probability of the happening of the event.
  • The probability of getting exactly ‘k’ successes in ‘n’ independent trials for a Random Variable X is expressed as P(X = k) and is given by the formula:

    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:

\(\rm ^6C_4\left(\dfrac23\right)^4\left(1-\dfrac23\right)^{6-4}+^6C_5\left(\dfrac23\right)^5\left(1-\dfrac23\right)^{6-5}+^6C_6\left(\dfrac23\right)^6\left(1-\dfrac23\right)^{6-6}\)

\(\rm (15)\left(\dfrac23\right)^4\left(\dfrac13\right)^2+(6)\left(\dfrac23\right)^5\left(\dfrac13\right)^1+(1)\left(\dfrac23\right)^6\left(\dfrac13\right)^0\)

\(\rm \dfrac{240}{729}+\dfrac{192}{729}+\dfrac{64}{729}\)

\(\rm \dfrac{496}{729}\).

 

  • nCr = \(\rm \dfrac {n!}{r!(n-r)!}\).
  • n! = 1 × 2 × 3 × ... × n.
  • 0! = 1.


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