Question

If on an average $1$ vessel in every $10$ is wrecked, the probability that out of $5$ vessels expected to arrive, $4$ at least will arrive safely is

• A. ${}_1^{5}C{\left(\frac{9}{10}\right)}^4\left(\frac{1}{10}\right){+}_0^{5}C{\left(\frac{9}{10}\right)}^5$
• B. $1{-}_1^{5}C{\left(\frac{9}{10}\right)}^4\left(\frac{1}{10}\right){-}_0^{5}C{\left(\frac{9}{10}\right)}^5$
• C. ${}_1^{5}C{\left(\frac{9}{10}\right)}^4\left(\frac{1}{10}\right)$
• D. ${}_0^{5}C{\left(\frac{9}{10}\right)}^5$

Answer 1

The correct option is A ${}_1^{5}C{\left(\frac{9}{10}\right)}^4\left(\frac{1}{10}\right){+}_0^{5}C{\left(\frac{9}{10}\right)}^5$

Probability of a vessel reacing safely is $0.9$.
The question asks for the probability of at least $4$ arriving safely. i.e. either 4 arrive or all the 5 arrive

$P\left(4\right)=$ $5C_4\times {\left(\frac{9}{10}\right)}^4\times \frac{1}{10}$
$P\left(5\right)$ = ${\left(\frac{9}{10}\right)}^5$
So the probability of at least $4$ reaching safely is $P\left(4\right)+P\left(5\right)$
= $5C_4\times {\left(\frac{9}{10}\right)}^4\times \frac{1}{10}$ + ${\left(\frac{9}{10}\right)}^5$ = Probability of a vessel reacing safely is $0.9$
The question asks for the probability of at least $4$ arriving safely. i.e. either $4$ arrive or all the $5$ arrive
$P\left(4\right)=$ $5C_4\times {\left(\frac{9}{10}\right)}^4\times \frac{1}{10}$
$P\left(5\right)=$ ${\left(\frac{9}{10}\right)}^5$
So the probability of at least $4$ reaching safely is $P\left(4\right)+P\left(5\right)$
= $5C_1\times {\left(\frac{9}{10}\right)}^4\times \frac{1}{10}$ + ${\left(\frac{9}{10}\right)}^5$