
In certain towns 40% of the people have brown hair,25% have brown eyes & 15% have both brown hair & brown eyes.If a person selected at random from the town has brown hair, the probability that he also has brown eyes is ?
$\begin{align}
& A)\dfrac{1}{5} \\
& B)\dfrac{3}{8} \\
& C)\dfrac{1}{3} \\
& D)\dfrac{2}{3} \\
\end{align}$
Answer
524.7k+ views
Hint: Represent probability of Brown Hair as \[P\left( BH \right)\]and Brown Eyes as \[P\left( BE \right).\] The people who have both brown hair and brown eyes probability is represented as intersection of BH and BE that is $P(BH\cap BE)$.When the person is selected at random and the probability that he also has brown eyes is determined by using conditional probability i.e
$P(\dfrac{BH}{BE})=\dfrac{P(BH\cap BE)}{P(BE)}$.
Complete step by step answer:
From the problem it is it given that ,
The brown hair people in a town =40%
Let us denote Brown Hair = BH
∴ The probability of brown hair people in a town =\[P\left( BH \right)\]=$\dfrac{40}{100}$=0.4⟶equation(1)
Similarly,
The brown eyes people in a town=25%
Let us denote Brown Eyes = BE
∴ The probability of brown eyes people in a town=\[P\left( BE \right)\]=$\dfrac{25}{100}$=2.5⟶equation(2)
Also given that the people who have both brown hair and brown eyes=15%
∴ The probability of the brown hair and brown eyes people = $P(BH\cap BE)$=$\dfrac{15}{100}$= 1.5⟶equation(3)
Now,
If a person is selected at random who has brown hair,the probability that he also has brown eyes is
will be found by using conditional probability.
We know that the conditional probability is represented as follows,
$P(\dfrac{BH}{BE})=\dfrac{P(BH\cap BE)}{P(BE)}$⟶equation(4)
Now substitute the equation(1),(2)&(3) in equation(4) we get,
$P(\dfrac{BH}{BE})=\dfrac{P(BH\cap BE)}{P(BE)}$
$P(\dfrac{BH}{BE})=\dfrac{0.15}{0.4}$
On simplifying the decimals to fractions we get as follows
$P(\dfrac{BH}{BE})=\dfrac{15}{40}$
On simplifying and dividing with “5” we get
$P(\dfrac{BH}{BE})=\dfrac{3}{8}$
So we have to choose now the appropriate option from the given problem.
Therefore ,the correct option is (B).
Note:
The students might go wrong here, “brown hair and brown eyes” represents that both are included so we have to take the intersection of both but not the union of the both.And this problem can also be solved easily by using the venn diagram representation method.
$P(\dfrac{BH}{BE})=\dfrac{P(BH\cap BE)}{P(BE)}$.
Complete step by step answer:
From the problem it is it given that ,
The brown hair people in a town =40%
Let us denote Brown Hair = BH
∴ The probability of brown hair people in a town =\[P\left( BH \right)\]=$\dfrac{40}{100}$=0.4⟶equation(1)
Similarly,
The brown eyes people in a town=25%
Let us denote Brown Eyes = BE
∴ The probability of brown eyes people in a town=\[P\left( BE \right)\]=$\dfrac{25}{100}$=2.5⟶equation(2)
Also given that the people who have both brown hair and brown eyes=15%
∴ The probability of the brown hair and brown eyes people = $P(BH\cap BE)$=$\dfrac{15}{100}$= 1.5⟶equation(3)
Now,
If a person is selected at random who has brown hair,the probability that he also has brown eyes is
will be found by using conditional probability.
We know that the conditional probability is represented as follows,
$P(\dfrac{BH}{BE})=\dfrac{P(BH\cap BE)}{P(BE)}$⟶equation(4)
Now substitute the equation(1),(2)&(3) in equation(4) we get,
$P(\dfrac{BH}{BE})=\dfrac{P(BH\cap BE)}{P(BE)}$
$P(\dfrac{BH}{BE})=\dfrac{0.15}{0.4}$
On simplifying the decimals to fractions we get as follows
$P(\dfrac{BH}{BE})=\dfrac{15}{40}$
On simplifying and dividing with “5” we get
$P(\dfrac{BH}{BE})=\dfrac{3}{8}$
So we have to choose now the appropriate option from the given problem.
Therefore ,the correct option is (B).
Note:
The students might go wrong here, “brown hair and brown eyes” represents that both are included so we have to take the intersection of both but not the union of the both.And this problem can also be solved easily by using the venn diagram representation method.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE
