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 .