Answer
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Hint: All the elements are independent to each other and are not connected in any way. So, we can directly multiply all the elements. Elements here are probabilities of choosing a girl, probability of choosing a rich person and probability of choosing a fair complexioned person, these three would consist probability of choosing a fair complexioned rich girl.
Complete step-by-step answer:
Probability is the quality or state of being probable or the extent to which something is likely to happen or be the case.
For example, there are 8 red balls and 4 green balls, then, the probability of choosing 1 red ball from these total ball would be:
\[\begin{align}
& \text{Denoted by }P=\dfrac{\text{a red ball (can be anyone among 8 red balls)}}{\text{Total number of balls}} \\
& P=\dfrac{8}{8+4}(\text{all 8 balls are alike)} \\
& \text{P=}\dfrac{8}{12}=\dfrac{2}{3} \\
\end{align}\]
Here, in the question, we are given a class which has 80 students, 25 of them are girls and 55 are boys, 10 students are rich and 20 students of them are fair complexioned. We have to find out the probability of selecting a fair complexioned rich girl.
This can also be written as:
P ( fair complexioned girl )
Now, P (fair complexioned girl) can also be viewed as first selecting a fair person from class, then selecting a person who is rich and then selecting a girl from the class.
Suppose,
P(F) = Probability of selecting a fair person from a class of 80 students.
P(R) = Probability of selecting a rich person from a class of 80 students.
P(G) = Probability of selecting a girl from a class of 80 students.
\[P(\text{fair complexioned girl)=P}\left( F\cap R\cap G \right)\]
$ \text{P}\left( F\cap R\cap G \right) $ it is the union of all the three probabilities F, R and G.
Union defines that, the given probability contains all the required criteria i.e. it will be a girl who is rich and fair complexioned.
Therefore,
\[\begin{align}
& \text{P}\left( F\cap R\cap G \right)=P(F)\times P(R)\times P(G) \\
& P(F)=\dfrac{20}{80}=\dfrac{1}{4} \\
& P(R)=\dfrac{10}{80}=\dfrac{1}{8} \\
& P(G)=\dfrac{25}{80}=\dfrac{5}{16} \\
& \therefore \text{P}\left( F\cap R\cap G \right)=\dfrac{1}{4}\times \dfrac{1}{8}\times \dfrac{5}{16} \\
& \Rightarrow \dfrac{5}{512} \\
\end{align}\]
So, the probability of choosing a fair complexioned girl is \[\dfrac{5}{512}\]
Note: Students must be clear while predicting the situation to be dependent or independent because in both the cases it would be different. After finding different probabilities students must not add them, they must be multiplied.
\[\begin{align}
& \text{P}\left( F\cap R\cap G \right)=\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{5}{16}\to \text{Wrong} \\
& \text{P}\left( F\cap R\cap G \right)=\dfrac{1}{4}\times \dfrac{1}{8}\times \dfrac{5}{16}\to Right \\
\end{align}\]
Complete step-by-step answer:
Probability is the quality or state of being probable or the extent to which something is likely to happen or be the case.
For example, there are 8 red balls and 4 green balls, then, the probability of choosing 1 red ball from these total ball would be:
\[\begin{align}
& \text{Denoted by }P=\dfrac{\text{a red ball (can be anyone among 8 red balls)}}{\text{Total number of balls}} \\
& P=\dfrac{8}{8+4}(\text{all 8 balls are alike)} \\
& \text{P=}\dfrac{8}{12}=\dfrac{2}{3} \\
\end{align}\]
Here, in the question, we are given a class which has 80 students, 25 of them are girls and 55 are boys, 10 students are rich and 20 students of them are fair complexioned. We have to find out the probability of selecting a fair complexioned rich girl.
This can also be written as:
P ( fair complexioned girl )
Now, P (fair complexioned girl) can also be viewed as first selecting a fair person from class, then selecting a person who is rich and then selecting a girl from the class.
Suppose,
P(F) = Probability of selecting a fair person from a class of 80 students.
P(R) = Probability of selecting a rich person from a class of 80 students.
P(G) = Probability of selecting a girl from a class of 80 students.
\[P(\text{fair complexioned girl)=P}\left( F\cap R\cap G \right)\]
$ \text{P}\left( F\cap R\cap G \right) $ it is the union of all the three probabilities F, R and G.
Union defines that, the given probability contains all the required criteria i.e. it will be a girl who is rich and fair complexioned.
Therefore,
\[\begin{align}
& \text{P}\left( F\cap R\cap G \right)=P(F)\times P(R)\times P(G) \\
& P(F)=\dfrac{20}{80}=\dfrac{1}{4} \\
& P(R)=\dfrac{10}{80}=\dfrac{1}{8} \\
& P(G)=\dfrac{25}{80}=\dfrac{5}{16} \\
& \therefore \text{P}\left( F\cap R\cap G \right)=\dfrac{1}{4}\times \dfrac{1}{8}\times \dfrac{5}{16} \\
& \Rightarrow \dfrac{5}{512} \\
\end{align}\]
So, the probability of choosing a fair complexioned girl is \[\dfrac{5}{512}\]
Note: Students must be clear while predicting the situation to be dependent or independent because in both the cases it would be different. After finding different probabilities students must not add them, they must be multiplied.
\[\begin{align}
& \text{P}\left( F\cap R\cap G \right)=\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{5}{16}\to \text{Wrong} \\
& \text{P}\left( F\cap R\cap G \right)=\dfrac{1}{4}\times \dfrac{1}{8}\times \dfrac{5}{16}\to Right \\
\end{align}\]
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