Question

In $n$ independent trials (finite) of a random experiment, let $X$ be the number of times an event $A$ occurs. If the probability of success of one trial say $p$ and we get the probability of failure of event $A$ i.e. $P\left(\overline{A}\right)=1-p=q$. The probability of $r$ success in $n$ trials is denoted by$P(X=r)$ such that $P(X=r){=}^n{C}_r{p}^r{q}^{n-r}$, known as binomial distribution of random variable $X$. We also have the following result-
(i) Probability of getting at least $k$ successes is
$P(X\ge K)={\sum }_{r=k}^n{}^n{C}_r{p}^r{q}^{n-r}$
(ii) Probability of getting at the most $k$ successes is
$P(X\le K)={\sum }_{r=0}^n{}^n{C}_r{p}^r{q}^{n-r}$
(iii) ${\sum }_{r=0}^n{=}^n{C}_r{p}^r{q}^{n-r}={(p+q)}^n=1$

On the basis of the above information answer the following question.

A product is supposed to contain 5% defective items. What is the probability that a sample of 8 items will contain less than 2 defeclive items?

• A. $\frac{25\times {\left(19\right)}^7}{{\left(20\right)}^8}$
• B. $\frac{27\times {\left(19\right)}^7}{{\left(20\right)}^8}$
• C. $\frac{{\left(19\right)}^7}{{\left(20\right)}^8}$
• D. None of these

Answer 1

Answer: (B)

$\frac{27\times {\left(19\right)}^7}{{\left(20\right)}^8}$

Here $n=8,p=\frac{5}{100}=\frac{1}{20}.q=\frac{19}{20}$

Now using $P(X=r)=C(8,r)p^r{q}^{8-r}$

$\therefore P(X=r<2)=P\left(0\right)+P\left(1\right)$

$={\left(\frac{19}{20}\right)}^8+\frac{8\times (19{)}^7}{(20{)}^8}=\frac{(19{)}^7}{(20{)}^8}[19+8]=\frac{27\times (19{)}^7}{(20{)}^8}$

Hence choice (b) is the correct answer.