Answer
Verified
471.9k+ views
Hint: Here, we will first assume that A be the event when Krishna will be alive and B be the event when Hari will be alive. Then we will use the property of probability, \[P\left( {{A^c}} \right) = 1 - P\left( A \right)\], where \[P\left( {{A^c}} \right)\] is complement of \[P\left( A \right)\]that is, Krishna will be dead and property of probability, \[P\left( {{B^c}} \right) = 1 - P\left( B \right)\], where \[P\left( {{B^c}} \right)\] is complement of \[P\left( B \right)\], that is, Hari will be dead. Then since both the events are independent of each other, we have \[P\left( {A \cap B} \right) = P\left( A \right)P\left( B \right)\] and getting \[P\left( {{A^c} \cap {B^c}} \right) = P\left( {{A^c}} \right)P\left( {{B^c}} \right)\] to find the required value.
Complete step-by-step answer:
Let us assume that A be the event when Krishna will be alive and B be the event when Hari will be alive.
So we given that the probabilities that Krishna will be alive 10 years hence, \[P\left( A \right) = \dfrac{7}{{15}}\] and that Hari will be alive is \[P\left( B \right) = \dfrac{7}{{10}}\].
Since we know that the property of probability, \[P\left( {{A^c}} \right) = 1 - P\left( A \right)\], where \[P\left( {{A^c}} \right)\] is complement of \[P\left( A \right)\]that is, Krishna will be dead, we get
\[
\Rightarrow P\left( {{A^c}} \right) = 1 - \dfrac{7}{{15}} \\
\Rightarrow P\left( {{A^c}} \right) = \dfrac{{15 - 7}}{{15}} \\
\Rightarrow P\left( {{A^c}} \right) = \dfrac{8}{{15}} \\
\]
Using the property of probability, \[P\left( {{B^c}} \right) = 1 - P\left( B \right)\], where \[P\left( {{B^c}} \right)\] is complement of \[P\left( B \right)\], that is, Hari will be dead, we get
\[
\Rightarrow P\left( {{B^c}} \right) = 1 - \dfrac{7}{{10}} \\
\Rightarrow P\left( {{B^c}} \right) = \dfrac{{10 - 7}}{{10}} \\
\Rightarrow P\left( {{B^c}} \right) = \dfrac{3}{{10}} \\
\]
Since both the events are independent of each other, we have \[P\left( {A \cap B} \right) = P\left( A \right)P\left( B \right)\].
So now substituting the values of \[P\left( {{A^c}} \right)\]and \[P\left( {{B^c}} \right)\]in the property of probability,\[P\left( {{A^c} \cap {B^c}} \right) = P\left( {{A^c}} \right)P\left( {{B^c}} \right)\], we get
\[
\Rightarrow P\left( {{A^c} \cap {B^c}} \right) = \dfrac{8}{{15}} \times \dfrac{3}{{10}} \\
\Rightarrow P\left( {{A^c} \cap {B^c}} \right) = \dfrac{{24}}{{150}} \\
\]
Hence, option A is correct.
Note: Students must know that they have to assume the given conditions with some distinct variables to understand the question properly. In solving these types of questions, students must know about basic properties of probability, \[P\left( {{A^c}} \right) = 1 - P\left( A \right)\] and since both the events are independent of each other, we have \[P\left( {A \cap B} \right) = P\left( A \right)P\left( B \right)\].
Complete step-by-step answer:
Let us assume that A be the event when Krishna will be alive and B be the event when Hari will be alive.
So we given that the probabilities that Krishna will be alive 10 years hence, \[P\left( A \right) = \dfrac{7}{{15}}\] and that Hari will be alive is \[P\left( B \right) = \dfrac{7}{{10}}\].
Since we know that the property of probability, \[P\left( {{A^c}} \right) = 1 - P\left( A \right)\], where \[P\left( {{A^c}} \right)\] is complement of \[P\left( A \right)\]that is, Krishna will be dead, we get
\[
\Rightarrow P\left( {{A^c}} \right) = 1 - \dfrac{7}{{15}} \\
\Rightarrow P\left( {{A^c}} \right) = \dfrac{{15 - 7}}{{15}} \\
\Rightarrow P\left( {{A^c}} \right) = \dfrac{8}{{15}} \\
\]
Using the property of probability, \[P\left( {{B^c}} \right) = 1 - P\left( B \right)\], where \[P\left( {{B^c}} \right)\] is complement of \[P\left( B \right)\], that is, Hari will be dead, we get
\[
\Rightarrow P\left( {{B^c}} \right) = 1 - \dfrac{7}{{10}} \\
\Rightarrow P\left( {{B^c}} \right) = \dfrac{{10 - 7}}{{10}} \\
\Rightarrow P\left( {{B^c}} \right) = \dfrac{3}{{10}} \\
\]
Since both the events are independent of each other, we have \[P\left( {A \cap B} \right) = P\left( A \right)P\left( B \right)\].
So now substituting the values of \[P\left( {{A^c}} \right)\]and \[P\left( {{B^c}} \right)\]in the property of probability,\[P\left( {{A^c} \cap {B^c}} \right) = P\left( {{A^c}} \right)P\left( {{B^c}} \right)\], we get
\[
\Rightarrow P\left( {{A^c} \cap {B^c}} \right) = \dfrac{8}{{15}} \times \dfrac{3}{{10}} \\
\Rightarrow P\left( {{A^c} \cap {B^c}} \right) = \dfrac{{24}}{{150}} \\
\]
Hence, option A is correct.
Note: Students must know that they have to assume the given conditions with some distinct variables to understand the question properly. In solving these types of questions, students must know about basic properties of probability, \[P\left( {{A^c}} \right) = 1 - P\left( A \right)\] and since both the events are independent of each other, we have \[P\left( {A \cap B} \right) = P\left( A \right)P\left( B \right)\].
Recently Updated Pages
A particle is undergoing a horizontal circle of radius class 11 physics CBSE
A particle is thrown vertically upwards with a velocity class 11 physics CBSE
A particle is rotated in a vertical circle by connecting class 11 physics CBSE
A particle is projected with a velocity v such that class 11 physics CBSE
A particle is projected with a velocity u making an class 11 physics CBSE
A particle is projected vertically upwards and it reaches class 11 physics CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Who was the leader of the Bolshevik Party A Leon Trotsky class 9 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Which is the largest saltwater lake in India A Chilika class 8 social science CBSE
Ghatikas during the period of Satavahanas were aHospitals class 6 social science CBSE