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# An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be atleast 4 successes.

Q. 9  An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be atleast 4 successes.

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Probability of success is twice the probability of failure.

Let probability of failure be X

then Probability of success = 2X

Sum of probabilities is 1.

$\therefore \, \, \, X+2X=1$

$\Rightarrow \, \, \, 3X=1$

$\Rightarrow \, \, \, X=\frac{1}{3}$

Let          $P=\frac{1}{3}$          and      $q=\frac{2}{3}$

Let X be random variable that represent the number of success in six trials.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X\geq 4)=P(X=4)+P(X=5)+P(X=6)$

$=^6C_4 \left [ \frac{2}{3} \right ]^4\left [ \frac{1}{3} \right ]^2+^6C_5 \left [ \frac{2}{3} \right ]^5\left [ \frac{1}{3} \right ]^1+^6C_6 \left [ \frac{2}{3} \right ]^6\left [ \frac{1}{3} \right ]^0$

$=\frac{15\times 2^4}{3^6}+\frac{6\times 2^5}{3^6}+\frac{2^6}{3^6}$

$=\frac{ 2^6}{3^6}(15+12+4)$

$=\frac{ 2^6}{3^6}(31)$

$=\frac{31}{9}\left [ \frac{2}{3} \right ]^4$

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