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In a trial, the probability of success is twice the probability of failure. In six trials, the probability of at most two failure will be
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Verified Answer
The correct answer is:
$\frac{496}{729}$
Let the probability of failure and success be $p$ and $q$, respectively.
Let $X$ represents the number of failure According to the question, $q=2 p$
$\because \quad p+q=1 \text { and } q=2 p$
$\therefore \quad p=\frac{1}{3} \text { and } q=\frac{2}{3}$
Now, required probability $=P(X \leq 2)$
$\begin{aligned}
& =P(X=0)+P(X=1)+P(X=2) \\
& ={ }^6 C_0 p^0 q^6+{ }^6 C_1 p^1 q^5+{ }^6 C_2 p^2 q^4 \\
& =\left(\frac{2}{3}\right)^6+6\left(\frac{1}{3}\right)^1\left(\frac{2}{3}\right)^5+15\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^4 \\
& =\frac{1}{729}(64+192+240)=\frac{496}{729}
\end{aligned}$
Let $X$ represents the number of failure According to the question, $q=2 p$
$\because \quad p+q=1 \text { and } q=2 p$
$\therefore \quad p=\frac{1}{3} \text { and } q=\frac{2}{3}$
Now, required probability $=P(X \leq 2)$
$\begin{aligned}
& =P(X=0)+P(X=1)+P(X=2) \\
& ={ }^6 C_0 p^0 q^6+{ }^6 C_1 p^1 q^5+{ }^6 C_2 p^2 q^4 \\
& =\left(\frac{2}{3}\right)^6+6\left(\frac{1}{3}\right)^1\left(\frac{2}{3}\right)^5+15\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^4 \\
& =\frac{1}{729}(64+192+240)=\frac{496}{729}
\end{aligned}$
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