Question

Three ships $A$ , $B$ and $C$ sail from England to India . If the ratio of their arriving safely are $2:5$ , $3:7$ and $6:11$ , respectively, then the probability of all the ships arriving safely is ?
A.$\dfrac{{18}}{{595}}$
B.$\dfrac{6}{{17}}$
C.$\dfrac{3}{{10}}$
D.$\dfrac{2}{7}$

Answer 1

Hint : The arrival of ships sailing from England to India are independent events , which means that the probability of one event occurring does not depend on the other . Therefore , the probability of the ship arriving $A$ does not depend on $B$ and $C$. Similarly , for $B$ and $C$ . Therefore , we will use the formula of probability of intersection of all the ships .

Complete step-by-step answer :
Let the probability of ship $A$ arriving India safely be $= P\left( A \right)$
Let the probability of ship $B$ arriving India safely be $= P\left( B \right)$
Let the probability of ship $C$ arriving India safely be $= P\left( C \right)$
The probability $P\left( A \right)$ will be $= \dfrac{2}{{2 + 5}}$
On solving we get ,
$P\left( A \right) = \dfrac{2}{7}$ .
Similarly , for the probability $P\left( B \right)$ we have
$P\left( B \right) = \dfrac{3}{{3 + 7}}$
On solving we get ,
$P\left( B \right) = \dfrac{3}{{10}}$
Also , for the probability $P\left( C \right) = \dfrac{6}{{6 + 11}}$
On solving we get ,
$P\left( C \right) = \dfrac{6}{{17}}$.
Now , for the probability of all the three ships arriving safely will be obtained by taking the intersection of the $A$, $B$ and $C$ which is given by $= P\left( {A \cap B \cap C} \right)$ , which is equals to
$P\left( {A \cap B \cap C} \right) = P\left( A \right) \times P\left( B \right) \times P\left( C \right)$ .
Now , putting the values of $P\left( A \right)$ , $P\left( B \right)$ and $P\left( C \right)$ we get ,
$P\left( {A \cap B \cap C} \right) = \dfrac{2}{7} \times \dfrac{3}{{10}} \times \dfrac{6}{{17}}$
On solving we get ,
$P\left( {A \cap B \cap C} \right) = \dfrac{{36}}{{1190}}$ ,
On simplifying we get ,
$P\left( {A \cap B \cap C} \right) = \dfrac{{18}}{{595}}$ .
Therefore , option (1) is the correct answer for this question .
So, the correct answer is “Option 1”.

Note : If there are $n$ elementary events associated with a random experiment and $m$of them are favorable to an event $A$ , then the probability of happening or occurrence of $A$ is denoted by $P\left( A \right)$ and is defined as ratio $\dfrac{m}{n}$ . The total probability of an event is equal to $1$ . Therefore , $P\left( A \right) + P\left( {\overline A } \right) = 1$ , where $P\left( {\overline A } \right)$ is the probability of non occurrence of the event .