Solveeit Logo
Login

Question

Question: In a group of 20 boys and 10 girls, 5 boys and 3 girls are honest. If two persons are selected at ra...

In a group of 20 boys and 10 girls, 5 boys and 3 girls are honest. If two persons are selected at random from the group, find the probability either both are girls or both are honest.

Explanation

Solution

First we will look at what is the probability of getting both the girls if two persons are selected and then we will look at the probability of selecting two honest persons from the group it doesn’t matter whether it’s a girl or boy, and then we will add these two probabilities to get the final answer.

Complete step-by-step solution
The probability of getting both the girls if two person are selected is let p1{{p}_{1}} ,
The formula that we are going to use to choose some persons from given amount of person is:
n!r!(n−r)!\dfrac{n!}{r!\left( n-r \right)!}
Therefore, p1{{p}_{1}}=
no. of ways of choosing two girlsno. of ways of choosing two person =10c230c2 =10!2!(10−2)!30!2!(30−2)! =10×930×29 \begin{aligned} & \dfrac{no.\text{ of ways of choosing two girls}}{no.\text{ of ways of choosing two person}} \\\ & =\dfrac{{}^{10}{{c}_{2}}}{{}^{30}{{c}_{2}}} \\\ & =\dfrac{\dfrac{10!}{2!\left( 10-2 \right)!}}{\dfrac{30!}{2!\left( 30-2 \right)!}} \\\ & =\dfrac{10\times 9}{30\times 29} \\\ \end{aligned}
The probability of getting both honest if two persons are selected is let p2{{p}_{2}},
Therefore, p2{{p}_{2}}=
no. of ways of choosing two honest personsno. of ways of choosing two persons\dfrac{no.\text{ of ways of choosing two honest persons}}{no.\text{ of ways of choosing two persons}}
Two honest persons can be chosen in three ways :
One girl and one boy, both boys, both girls.
So now we will find the value of p2{{p}_{2}},
=5c2+3c2+5c13c130c2 =5!2!(5−2)!+3!2!(3−2)!+5!(5−1)!.3!(3−1)!30!2!(30−2)! =10+3+1515×29 =2815×29 \begin{aligned} & =\dfrac{{}^{5}{{c}_{2}}+{}^{3}{{c}_{2}}+{}^{5}{{c}_{1}}{}^{3}{{c}_{1}}}{{}^{30}{{c}_{2}}} \\\ & =\dfrac{\dfrac{5!}{2!\left( 5-2 \right)!}+\dfrac{3!}{2!\left( 3-2 \right)!}+\dfrac{5!}{\left( 5-1 \right)!}.\dfrac{3!}{\left( 3-1 \right)!}}{\dfrac{30!}{2!\left( 30-2 \right)!}} \\\ & =\dfrac{10+3+15}{15\times 29} \\\ & =\dfrac{28}{15\times 29} \\\ \end{aligned}
Now the final answer will be,
p2{{p}_{2}} + p1{{p}_{1}}
=2815×29+4515×29 =7315×29 =73435 \begin{aligned} & =\dfrac{28}{15\times 29}+\dfrac{45}{15\times 29} \\\ & =\dfrac{73}{15\times 29} \\\ & =\dfrac{73}{435} \\\ \end{aligned}
Hence, this is the final answer of probability either both are girls or both are honest.

Note: One can also solve this question by manually choosing two girls from the group of people and choosing two honest people from the group of people, and then we have to find the total number of cases by counting manually and then we will get the desired answer. The possible mistake here is that not considering three cases for choosing two honest persons from the given group.