
$ 60 $ percent people in a group of $ 10 $ people, have brown eyes. Two people are to be selected at random from the group. What is the probability that neither person selected will have brown eyes?
A) 0.13
B) 0.18
C) 0.25
D) 0.36
Answer
578.1k+ views
Hint: To find a solution we first find the number of people with brown eyes from a given percentage and using this we find the number of people who don’t have brown eyes using total number of people. Then, to find the required probability we find the ratio of the number of ways of selecting two persons out of a person who don’t have brown eyes to the total number of ways of selecting two people out of ten.
Probability of any event is given as: $ \dfrac{{number\,\,of\,\,favourable\,\,outcomes}}{{totla\,\,number\,\,of\,\,outcomes}} $
Formula for combination: $ ^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $
Complete step-by-step answer:
Total number of people in a group = $ 10 $
Number of people having brown eye = $ 60\% $
$
\Rightarrow \,\,number\,\,of\,\,people\,\,with\,\,brown\,\,eye = 60\% \left( {10} \right) \\
\Rightarrow \,\,number\,\,of\,\,people\,\,with\,\,brown\,\,eye = \dfrac{{60}}{{100}} \times \left( {10} \right) \\
\Rightarrow \,\,number\,\,of\,\,people\,\,with\,\,brown\,\,eye = 6 \;
$
Therefore, out of $ 10 $ persons there are $ 6 $ persons with brown eye.
Number of people who don’t have brown eye = $ 10 - 6 = 4 $
Number of way in which we can select two person who don’t have brown eye given as: $ ^4{C_2} $
= $ \dfrac{{4!}}{{2!2!}} $
$
\Rightarrow \dfrac{{4 \times 3}}{2} \\
= 6 \;
$
Number of way in which we can select two person out of ten people given as: $ ^{10}{C_2} $
= $ \dfrac{{10!}}{{2!8!}} $
$
\Rightarrow \dfrac{{10 \times 9}}{2} \\
= 45 \;
$
Therefore, probability of selecting two person out of ten who don’t have brown eye = $ \dfrac{6}{{45}} $
= $ 0.13 $
So, the correct answer is “Option A”.
Note: We can also find the solution of the problem in another way. In this we calculate the total probabilities of selecting a person with brown eyes and then subtracting sub of probability so obtain from $ 1 $ to get probability that neither of a person selected will have brown eyes.
Probability of any event is given as: $ \dfrac{{number\,\,of\,\,favourable\,\,outcomes}}{{totla\,\,number\,\,of\,\,outcomes}} $
Formula for combination: $ ^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $
Complete step-by-step answer:
Total number of people in a group = $ 10 $
Number of people having brown eye = $ 60\% $
$
\Rightarrow \,\,number\,\,of\,\,people\,\,with\,\,brown\,\,eye = 60\% \left( {10} \right) \\
\Rightarrow \,\,number\,\,of\,\,people\,\,with\,\,brown\,\,eye = \dfrac{{60}}{{100}} \times \left( {10} \right) \\
\Rightarrow \,\,number\,\,of\,\,people\,\,with\,\,brown\,\,eye = 6 \;
$
Therefore, out of $ 10 $ persons there are $ 6 $ persons with brown eye.
Number of people who don’t have brown eye = $ 10 - 6 = 4 $
Number of way in which we can select two person who don’t have brown eye given as: $ ^4{C_2} $
= $ \dfrac{{4!}}{{2!2!}} $
$
\Rightarrow \dfrac{{4 \times 3}}{2} \\
= 6 \;
$
Number of way in which we can select two person out of ten people given as: $ ^{10}{C_2} $
= $ \dfrac{{10!}}{{2!8!}} $
$
\Rightarrow \dfrac{{10 \times 9}}{2} \\
= 45 \;
$
Therefore, probability of selecting two person out of ten who don’t have brown eye = $ \dfrac{6}{{45}} $
= $ 0.13 $
So, the correct answer is “Option A”.
Note: We can also find the solution of the problem in another way. In this we calculate the total probabilities of selecting a person with brown eyes and then subtracting sub of probability so obtain from $ 1 $ to get probability that neither of a person selected will have brown eyes.
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