Question

On an average nine out of $10$ ships that have departed at $A$ reach $B$ safely. The probability that out of five ships that have departed at $A$ at least four will reach $B$ safely is

• A. $14{\left(0.9\right)}^5$
• B. $1.4{\left(0.9\right)}^5$
• C. $0.14{\left(0.9\right)}^4$
• D. $1.4{\left(0.9\right)}^4$

Answer 1

The correct option is D $1.4{\left(0.9\right)}^4$

Let the probability of a ship reaching $B$ safely be $p$ and the probability of not reaching safely be $q$.

Then $p=0.9,q=1-p=0.1$

We need the probability of atleast $4$ ships reaching safely,i.e,

$\Rightarrow$ $P(x=4)+P(x=5)$
$={}^5{C}_1\left(0.1\right)\times {\left(0.9\right)}^4+{\left(0.9\right)}^5=\left(1.4\right){\left(0.9\right)}^4$