Aspire Faculty ID #10359 · Topic: NIMCET 2013 · Just now
NIMCET 2013

An experiment succeeds twice often as it fails. The probability that in the next six trials there will be at least four successes is.

Solution

 An experiment succeeds twice as often as it fails. 
 Let success probability = $p$ 
 Then failure probability = $1-p$ 
 Given: $p = 2(1-p)$ $\Rightarrow p = 2 - 2p$ $\Rightarrow 3p = 2$ $\Rightarrow p = \frac{2}{3}$ 
 So failure probability = $\frac{1}{3}$. 
 We perform 6 independent trials. 
 Let $X$ = number of successes. 
 We need: $P(X \ge 4)$. 
 $P(X \ge 4) = P(X=4) + P(X=5) + P(X=6)$ 
 Use binomial formula: $P(X=k) = {6 \choose k} \left(\frac{2}{3}\right)^k \left(\frac{1}{3}\right)^{6-k}$ 

 1. $P(X=4)$ $= {6 \choose 4} \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^2$ $= 15 \cdot \frac{16}{81} \cdot \frac{1}{9}$ $= 15 \cdot \frac{16}{729}$ $= \frac{240}{729}$ 

 2. $P(X=5)$ $= {6 \choose 5} \left(\frac{2}{3}\right)^5 \left(\frac{1}{3}\right)$ $= 6 \cdot \frac{32}{243} \cdot \frac{1}{3}$ $= 6 \cdot \frac{32}{729}$ $= \frac{192}{729}$ 

 3. $P(X=6)$ $= {6 \choose 6} \left(\frac{2}{3}\right)^6$ $= 1 \cdot \frac{64}{729}$ $= \frac{64}{729}$ 

 Add them: 
 $P(X \ge 4) = \frac{240}{729} + \frac{192}{729} + \frac{64}{729}$ 
 $= \frac{496}{729}$ 
 Final Answer: $\frac{496}{729}$
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