A class consists of \[100\] students, \[25\] of them are girls and \[75\] are boys, \[20\] of them are rich and remaining poor, \[40\] of them are fair complexioned. The probability of selecting a fair complexioned rich girl is
.
Last updated date: 15th Mar 2023
•
Total views: 303.9k
•
Views today: 6.83k
Answer
303.9k+ views
Hint: Calculate the probability of each of the given events and use the formula for calculating the probability of intersection of independent events, which states that the probability of intersection of independent events is the product of probability of each of the individual events.
Complete step-by-step answer:
We have the data regarding a class of \[100\] students. We have to find the probability of selecting a fair complexioned rich girl.
Let’s denote the event of getting a girl by \[A\] and thus, the event of getting a boy is denoted by \[{{A}^{c}}\].
Similarly, denote the event of getting a rich person by \[B\] and thus, the event of getting a poor person is denoted by \[{{B}^{c}}\].
Similarly, the event of getting a fair complexioned person is denoted by \[C\] and thus, the event of getting a dark complexioned person is denoted by \[{{C}^{c}}\].
We have to calculate the probability of each of these events. We know that the probability of getting any event is the ratio of the number of favourable outcomes to the total number of possible outcomes.
As there are \[25\] girls in a class, we have \[P\left( A \right)=\dfrac{25}{100}=\dfrac{1}{4}\] and \[P\left( {{A}^{c}} \right)=\dfrac{75}{100}=\dfrac{3}{4}\].
Similarly, as we have \[20\] rich students in a class, we have \[P\left( B \right)=\dfrac{20}{100}=\dfrac{1}{5}\] and \[P\left( {{B}^{c}} \right)=\dfrac{80}{100}=\dfrac{4}{5}\].
As there are \[40\] fair complexioned students in a class, we have \[P\left( C \right)=\dfrac{40}{100}=\dfrac{2}{5}\] and \[P\left( {{C}^{c}} \right)=\dfrac{60}{100}=\dfrac{3}{5}\].
We now have to evaluate the probability of getting a fair complexioned rich girl, which is \[P\left( A\cap B\cap C \right)\]. We observe that the events \[A,B,C\] are independent, which means that the occurrence or non-occurrence of one event doesn’t affect the occurrence or non-occurrence of another event.
We know that if events \[A,B,C\] are independent, we have \[P\left( A\cap B\cap C \right)=P\left( A \right)P\left( B \right)P\left( C \right)\].
Substituting the values from above equations, we have \[P\left( A\cap B\cap C \right)=P\left( A \right)P\left( B \right)P\left( C \right)=\dfrac{1}{4}\times \dfrac{1}{5}\times \dfrac{2}{5}=\dfrac{1}{50}\].
Hence, the probability of getting a fair complexioned rich girl, denoted by \[P\left( A\cap B\cap C \right)\], is equal to \[\dfrac{1}{50}\].
Note: Probability of any event describes how likely an event is to occur or how likely it is that a proposition is true. The value of probability of any event always lies in the range \[\left[ 0,1 \right]\] where having \[0\] probability indicates that the event is impossible to happen, while having probability equal to \[1\] indicates that the event will surely happen. We must remember that the sum of probability of occurrence of some event and probability of non-occurrence of the same event is always \[1\].
Complete step-by-step answer:
We have the data regarding a class of \[100\] students. We have to find the probability of selecting a fair complexioned rich girl.
Let’s denote the event of getting a girl by \[A\] and thus, the event of getting a boy is denoted by \[{{A}^{c}}\].
Similarly, denote the event of getting a rich person by \[B\] and thus, the event of getting a poor person is denoted by \[{{B}^{c}}\].
Similarly, the event of getting a fair complexioned person is denoted by \[C\] and thus, the event of getting a dark complexioned person is denoted by \[{{C}^{c}}\].
We have to calculate the probability of each of these events. We know that the probability of getting any event is the ratio of the number of favourable outcomes to the total number of possible outcomes.
As there are \[25\] girls in a class, we have \[P\left( A \right)=\dfrac{25}{100}=\dfrac{1}{4}\] and \[P\left( {{A}^{c}} \right)=\dfrac{75}{100}=\dfrac{3}{4}\].
Similarly, as we have \[20\] rich students in a class, we have \[P\left( B \right)=\dfrac{20}{100}=\dfrac{1}{5}\] and \[P\left( {{B}^{c}} \right)=\dfrac{80}{100}=\dfrac{4}{5}\].
As there are \[40\] fair complexioned students in a class, we have \[P\left( C \right)=\dfrac{40}{100}=\dfrac{2}{5}\] and \[P\left( {{C}^{c}} \right)=\dfrac{60}{100}=\dfrac{3}{5}\].
We now have to evaluate the probability of getting a fair complexioned rich girl, which is \[P\left( A\cap B\cap C \right)\]. We observe that the events \[A,B,C\] are independent, which means that the occurrence or non-occurrence of one event doesn’t affect the occurrence or non-occurrence of another event.
We know that if events \[A,B,C\] are independent, we have \[P\left( A\cap B\cap C \right)=P\left( A \right)P\left( B \right)P\left( C \right)\].
Substituting the values from above equations, we have \[P\left( A\cap B\cap C \right)=P\left( A \right)P\left( B \right)P\left( C \right)=\dfrac{1}{4}\times \dfrac{1}{5}\times \dfrac{2}{5}=\dfrac{1}{50}\].
Hence, the probability of getting a fair complexioned rich girl, denoted by \[P\left( A\cap B\cap C \right)\], is equal to \[\dfrac{1}{50}\].
Note: Probability of any event describes how likely an event is to occur or how likely it is that a proposition is true. The value of probability of any event always lies in the range \[\left[ 0,1 \right]\] where having \[0\] probability indicates that the event is impossible to happen, while having probability equal to \[1\] indicates that the event will surely happen. We must remember that the sum of probability of occurrence of some event and probability of non-occurrence of the same event is always \[1\].
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
