Question
In a multiple-choice examination with three possible answers for each of the five questions out of which only one is correct, what is the probability that a candidate would get four or more correct answers just by guessing?
Probability of getting a correct answer is,
$\text{p}=\frac{1}{3}$$\therefore\text{q}=1-\text{p}=1-\frac{1}{3}=\frac{2}{3}$
Clearly, X has a binomial distribution with
$\text{n = 5}$ and $\text{p}=\frac{1}{3}$$\therefore\text{P(X = x)}=\text{ }^{\text{n}}\text{C}_{\text{x}}\text{q}^{\text{n}-\text{x}}\text{p}^{\text{x}}$
$=\text{ }^5\text{C}_{\text{x}}\big(\frac{2}{3}\big)^{5-\text{x}}.\big(\frac{1}{3}\big)^{\text{x}}$
P(guessing more than 4 correct answer)
$=\text{P(X}\geq4)$$=\text{P(X}=4)+\text{P(X}=5)$
$=\text{ }^5\text{C}_4\big(\frac{2}{3}\big).\big(\frac{1}{3}\big)^4+\text{ }^5\text{C}_5\big(\frac{1}{3}\big)^5$
$=5.\frac{2}{3}.\frac{1}{81}+1.\frac{1}{243}$
$=\frac{10}{243}+\frac{1}{243}$
$=\frac{11}{243}$
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$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$ and $\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\big)+\mu\big(-\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$