Q101 / 180 · COMEDK 2023 (Memory Based) · Mathematics

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

  • A
    $\frac{600}{729}$
  • B
    $\frac{500}{729}$
  • C
    $\frac{400}{729}$
  • Correct answer
    D
    $\frac{496}{729}$

Solution

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 $=P(X \leq 2)$ $\begin{aligned} & =P(X=0)+P(X=1)+P(X=2) \\ & ={ }^6 C_0 p^0 q^6+{ }^6 C_1 p^1 q^5+{ }^6 C_2 p^2 q^4 \\ & =\left(\frac{2}{3}\right)^6+6\left(\frac{1}{3}\right)^1\left(\frac{2}{3}\right)^5+15\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^4 \\ & =\frac{1}{729}(64+192+240)=\frac{496}{729} \end{aligned}$

Hence, the correct answer is option: 'D'.

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